Looking for malloc() help

[SP]

I am learning C and have a question re: malloc().

I wrote simple program which assigns a value to a structure and then
prints it as follow:

#include <stdio.h>
#include <stdlib.h>

struct item {
char name[20];
int quantity;
};

int main (int argc, char *argv[])
{
struct item *stuff;

//allocate memory for structure
stuff = malloc(3 * sizeof(struct item));

strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;

printf("%s %d\n", stuff[1].name, st.quantity);
printf("%s %d\n", stuff[2].name, st.quantity);

free(stuff);
return 0;
}

I then change the structure declaration in order to dynamically
allocate memory
for the char array:
struct item {
char *name;
int quantity;
};

The program compiles with no errors, but it crashes with the following
error:

line 3: 2485 Segmentation fault

Thanks for your help

 

[xiaohuamao]

I don't know what is the problem. The following code works in my
machine.
struct item {
char* name;
int quantity;
};

int main ()
{
struct item *stuff;

//allocate memory for structure
stuff = (struct item *)malloc(3 * sizeof(struct item));

stuff[1].name = (char *) malloc(sizeof("apple"));
strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
stuff[2].name = (char *) malloc(sizeof("banana"));
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;


printf("%s %d\n", stuff[1].name, stuff[1].quantity);
printf("%s %d\n", stuff[2].name, stuff[2].quantity);


free(stuff);
return 0;
}

 

[SP]

I don't know what is the problem. The following code works in my
machine.
struct item {
char* name;
int quantity;
};

int main ()
{
struct item *stuff;

//allocate memory for structure
stuff = (struct item *)malloc(3 * sizeof(struct item));

stuff[1].name = (char *) malloc(sizeof("apple"));
strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
stuff[2].name = (char *) malloc(sizeof("banana"));
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;


printf("%s %d\n", stuff[1].name, stuff[1].quantity);
printf("%s %d\n", stuff[2].name, stuff[2].quantity);


free(stuff);
return 0;
}

Not sure it matters, my PC is as follows:

Linux Slackware 10.1
compiling program with "gcc -o test struct.c"

Is anyone else able to run this ?

 

[Thomas Lumley]

I am learning C and have a question re: malloc().

I wrote simple program which assigns a value to a structure and then
prints it as follow:

#include <stdio.h>
#include <stdlib.h>

struct item {
char name[20];
int quantity;
};

int main (int argc, char *argv[])
{
struct item *stuff;

//allocate memory for structure
stuff = malloc(3 * sizeof(struct item));

strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;

printf("%s %d\n", stuff[1].name, st.quantity);

st has not been defined. You probably mean stuff[1].quantity

printf("%s %d\n", stuff[2].name, st.quantity);
stuff[2].quantity

free(stuff);
return 0;
}
I then change the structure declaration in order to dynamically
allocate memory
for the char array:
struct item {
char *name;
int quantity;
};

The program compiles with no errors, but it crashes with the following
error:

In the first case the struct contains memory (20 bytes) for storing the
names so the malloc() for the structs is sufficient. In the second
case the struct contains only a pointer, so you are copying "apple" to
whatever random address ends up in stuff[1].name. Undefined behaviour.
[*]

You could either use malloc() to allocate space for storing the name
and assign the resulting pointer to stuff[1].name, or just do
stuff[1].name = "apple";
and not copy the string at all.

In this case just assigning the pointer would make more sense, but in a
real example either strategy might be appropriate.

-thomas


[*] Technically the undefined behaviour could be even worse than
copying to some random address, but that's bad enough.

 

[Keith Thompson]

I am learning C and have a question re: malloc().

I wrote simple program which assigns a value to a structure and then
prints it as follow:

#include <stdio.h>
#include <stdlib.h>

struct item {
char name[20];
int quantity;
};

int main (int argc, char *argv[])
{
struct item *stuff;

//allocate memory for structure
stuff = malloc(3 * sizeof(struct item));

strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;

printf("%s %d\n", stuff[1].name, st.quantity);
printf("%s %d\n", stuff[2].name, st.quantity);

free(stuff);
return 0;
}

I then change the structure declaration in order to dynamically
allocate memory
for the char array:
struct item {
char *name;
int quantity;
};

The program compiles with no errors, but it crashes with the following
error:

line 3: 2485 Segmentation fault

So you've shown us the program that works, but not the one that
doesn't.

Ok, I'll assume that the change in the declaration of "struct item"
was the only change you made. The problem is that you haven't
allocated space for the strings "apple" and "banana". You need
to do something like:

stuff[1].name = malloc(some_number_of_bytes);
strcpy(stuff[1].name, "apple");

Some other notes:

If you don't use argc and argv, you can omit their declarations:
int main(void)

Always check the result of malloc(). If it fails for whatever reason,
you go on and try to access the memory anyway; this can Make Bad
Things Happen. Even if you just abort the program on failure, it's
better than ignoring it.

Your malloc() call is good, but there's a little trick that can make
it even better:

stuff = malloc(3 * sizeof *stuff);

This way, you don't have to repeat (and possibly get wrong) the type
of "stuff" in the malloc call, and if you change the type of stuff you
don't have to track down and change all the calls.

I presume you know you're using only the 2nd and 3rd of the 3 elements
you allocated for your array.

It's generally considered a good idea to avoid "//" comments when
posting to Usenet. News software often wraps long lines. If a "//"
comment is wrapped, it usually creates a syntax error; if a "/*
.... */" comment is wrapped, it's usually harmless. (And "//" comments
aren't supported in C90.)

 

[Richard Heathfield]

SP said:

I am learning C and have a question re: malloc().

I wrote simple program which assigns a value to a structure and then
prints it as follow:

#include <stdio.h>
#include <stdlib.h>

struct item {
char name[20];
int quantity;
};

int main (int argc, char *argv[])
{
struct item *stuff;

//allocate memory for structure
stuff = malloc(3 * sizeof(struct item));

Not bad, but better would be:

stuff = malloc(3 * sizeof *stuff); /* note the * in front of stuff */

because there's less to get wrong (you don't need to type the typename, so
you can't mess it up - not that you did, on this occasion), and it's more
robust in the face of subsequent changes to the pointer type.

But malloc can fail. In that eventuality, it will return NULL. You should
check for this before relying on the allocation.

For a "student exercise" such as this one, the following would be
sufficient:

if(stuff == NULL)
{
fputs("Insufficient memory to continue. Terminating program.\n",
stderr);
exit(EXIT_FAILURE);
}

strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;

You're not using stuff[0], then? Well, okay, it's your memory...

strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;

printf("%s %d\n", stuff[1].name, st.quantity);
printf("%s %d\n", stuff[2].name, st.quantity);

free(stuff);
return 0;
}

That's fine, apart from what I've pointed out already.

I then change the structure declaration in order to dynamically
allocate memory
for the char array:
struct item {
char *name;
int quantity;
};

The program compiles with no errors, but it crashes with the following
error:

line 3: 2485 Segmentation fault

Yes. Now that you've changed name from being an array to being a mere
pointer, you need to point it at some storage.

if(stuff != NULL)
{
int i;
for(i = 0; i < 3; i++)
{
stuff.name = malloc(20 * sizeof *stuff.name);
if(stuff.name == NULL)
{
fprintf(stderr, "Can't allocate space for name %d\n", i);
fprintf(stderr, "Terminating program.\n");
exit(EXIT_FAILURE);
}
}

You will now be able to do your strcpy in safety (provided the string you
copy into it is 19 or fewer bytes long).

Once you've finished, and before you free(stuff), you should do this:

for(i = 0; i < 3; i++)
{
free(stuff.name);
}

Then you can free(stuff);

 

[Richard Heathfield]

Thomas Lumley said:

SP wrote:

printf("%s %d\n", stuff[1].name, st.quantity);

st has not been defined. You probably mean stuff[1].quantity

printf("%s %d\n", stuff[2].name, st.quantity);

Well spotted. I missed those completely.

 

[SP]

Thomas Lumley said:

SP wrote:

printf("%s %d\n", stuff[1].name, st.quantity);

st has not been defined. You probably mean stuff[1].quantity

printf("%s %d\n", stuff[2].name, st.quantity);

Well spotted. I missed those completely.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Allocating memory for the char array was the part I didnt know how to
do,
now it works just fine.

Thanks to all.

 

[Herbert Rosenau]

I don't know what is the problem. The following code works in my
machine.
struct item {
char* name;
int quantity;
};

int main ()
{
struct item *stuff;

//allocate memory for structure
stuff = (struct item *)malloc(3 * sizeof(struct item));

Never ever cast the result from malloc() except you like to go into
the lands of undefined behavior and you dont like to get diagnostics
from your compiler when you've forget to include stdlib.h.

Never ever cast something - excet you knows exactly what you're
doing. casting the result of a function returning void* is playing
with the health of y<or app because you does NOT know what you does.
When you would know what you does you would never cast that.

stuff[1].name = (char *) malloc(sizeof("apple"));

see above

strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
stuff[2].name = (char *) malloc(sizeof("banana"));

see aboove

strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;


printf("%s %d\n", stuff[1].name, stuff[1].quantity);
printf("%s %d\n", stuff[2].name, stuff[2].quantity);


free(stuff);

Before you frees stuff you should free its members you've allocated
with malloc() to avoid memory leaks.

return 0;
}

Array starts with index 0, not 1. Why does you malloc 3 struct members
when you only needs 2? And why does you use the last but not the first
ones?

--
Tschau/Bye
Herbert

Visit http://www.ecomstation.de the home of german eComStation
eComStation 1.2 Deutsch ist da!

 

[Herbert Rosenau]

Thomas Lumley said:

SP wrote:

printf("%s %d\n", stuff[1].name, st.quantity);
st has not been defined. You probably mean stuff[1].quantity
printf("%s %d\n", stuff[2].name, st.quantity);

Well spotted. I missed those completely.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Allocating memory for the char array was the part I didnt know how to
do,
now it works just fine.

No, you've forgotten to include stdlib.h so even as it seems to work
you're sitting in the middle of the lands of undefined behavior.

Thanks to all.

--
Tschau/Bye
Herbert

Visit http://www.ecomstation.de the home of german eComStation
eComStation 1.2 Deutsch ist da!

 

[Guest]

Never ever cast the result from malloc() except you like to go into
the lands of undefined behavior and you dont like to get diagnostics
from your compiler when you've forget to include stdlib.h.

Or if you want C++ compatibility, for whatever reason. (It may not make
sense for most C code, but one of the times it does make sense is if
it's an inline function in a header file that's kept after
installation.)

Never ever cast something - excet you knows exactly what you're
doing. casting the result of a function returning void* is playing
with the health of y<or app because you does NOT know what you does.
When you would know what you does you would never cast that.

You can quite legitimately cast a void * return value if you want to
store it as an (u)intptr_t value.

I wouldn't have commented if you hadn't so specifically said "never".

 

[Joe Wright]

I am learning C and have a question re: malloc().

I wrote simple program which assigns a value to a structure and then
prints it as follow:

#include <stdio.h>
#include <stdlib.h>

struct item {
char name[20];
int quantity;
};

int main (int argc, char *argv[])
{
struct item *stuff;

//allocate memory for structure
stuff = malloc(3 * sizeof(struct item));

strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;

printf("%s %d\n", stuff[1].name, st.quantity);
printf("%s %d\n", stuff[2].name, st.quantity);

free(stuff);
return 0;
}

I then change the structure declaration in order to dynamically
allocate memory
for the char array:
struct item {
char *name;
int quantity;
};

The program compiles with no errors, but it crashes with the following
error:

line 3: 2485 Segmentation fault

Thanks for your help

But you didn't allocate the memory. You'll need something like..

stuff[1].name = malloc(20);

...before

strcpy(stuff[1].name, "apple");

...and then you must free stuff[1].name (and stuff[2].name) before
freeing stuff.

 

[Herbert Rosenau]

Herbert Rosenau wrote:

Or if you want C++ compatibility, for whatever reason. (It may not make
sense for most C code, but one of the times it does make sense is if
it's an inline function in a header file that's kept after
installation.)

No. In C++ you would use simply operator new.

You can quite legitimately cast a void * return value if you want to
store it as an (u)intptr_t value.

You can burn your home, no question. Bit is it a good idea to do so?
You can kill your wife too. Casting is insidious and casing void* to
anything else says nothing but you knows not how to handle pointers to
void.

I wouldn't have commented if you hadn't so specifically said "never".

It is simply: never ever cast the result of a function returning a
pointer to void. You'll end up in the lands of undifined behavior..

--
Tschau/Bye
Herbert

Visit http://www.ecomstation.de the home of german eComStation
eComStation 1.2 Deutsch ist da!

 

[Guest]

No. In C++ you would use simply operator new.

new is not an option when writing header files that work both for C and
for C++ code.

You can burn your home, no question. Bit is it a good idea to do so?
You can kill your wife too. Casting is insidious and casing void* to
anything else says nothing but you knows not how to handle pointers to
void.

How would you use (u)intptr_t, then? (If you say you wouldn't at all,
keep in mind that it's a type defined in standard C(99).)

It is simply: never ever cast the result of a function returning a
pointer to void. You'll end up in the lands of undifined behavior..

That's clearly untrue, and depending on your intent quite possibly a
lie. There's no undefined behaviour in casting the result of malloc().
It's usually poor style, and it can cover up warnings when the
behaviour would be just as undefined without a cast, but that's all.

 

[Herbert Rosenau]

new is not an option when writing header files that work both for C and
for C++ code.

But you says that it is good to write programs for both C anf FORTRAN
or C++ and Cobol?


You can't write a single program that uses two highly different
languages. When you needs to write failsave programs instead of
hacking blind around you have to use either C or C++. Mixing up
languages inside the same program ends always up with holes of any
kind.

How would you use (u)intptr_t, then? (If you say you wouldn't at all,
keep in mind that it's a type defined in standard C(99).)

I would use it as it is defined. Conversion between all kinds of data
pointer to/from pointer to void is done already witrhout cast. When
you does not know what you does you should avoid cats anyway.

That's clearly untrue, and depending on your intent quite possibly a
lie. There's no undefined behaviour in casting the result of malloc().
It's usually poor style, and it can cover up warnings when the
behaviour would be just as undefined without a cast, but that's all.

You must learn C before you can use the language, so start learning C.
Until you have learned C you are absolutely unable to something about
the language. You've proven now that you knows nothing about C.

Will you guarantee that in each and any environment a function
returning a pointer uses always the same location as a function
returning int, even as the standard since more than 15 years says the
contrary.

You knows nothing about C but tells nonsense. There are more traps you
falls in as you can dream of when you does not understund what C is -
and you have not even basic knowlede about C yet.

--
Tschau/Bye
Herbert

Visit http://www.ecomstation.de the home of german eComStation
eComStation 1.2 Deutsch ist da!

 

[Keith Thompson]

But you says that it is good to write programs for both C anf FORTRAN
or C++ and Cobol?

No, he didn't say that.

You can't write a single program that uses two highly different
languages. When you needs to write failsave programs instead of
hacking blind around you have to use either C or C++. Mixing up
languages inside the same program ends always up with holes of any
kind.

C and C++ are not "highly different"; C is nearly (but not exactly) a
subset of C++, and it's possible to write reasonable code within the
intersection of the two languages.

It rarely makes sense to do so. 99% of the time, it makes more sense
to write in pure C, or to write in pure C++, or to use C++'s 'extern
"C"' mechanism if you need to use both. But it's also possible to
write library code that's intended to be used by either C or C++
client code. P.J. Plauger does this, for example.

 

[Richard Heathfield]

Keith Thompson said:

But it's also possible to
write library code that's intended to be used by either C or C++
client code. P.J. Plauger does this, for example.

And the reason his name is always given as an example is that he's almost
the only guy in the universe with a good reason to do this.

 

[Guest]

I would use it as it is defined. Conversion between all kinds of data
pointer to/from pointer to void is done already witrhout cast. When
you does not know what you does you should avoid cats anyway.

It is impossible in standard C to store a pointer value in an
(u)intptr_t object without a cast. Using a cast in this case, according
to you, shows you don't know how to handle pointers. So I ask again,
how would you use (u)intptr_t?

That's clearly untrue, and depending on your intent quite possibly a
lie. There's no undefined behaviour in casting the result of malloc().
It's usually poor style, and it can cover up warnings when the
behaviour would be just as undefined without a cast, but that's all.

[Insult snipped]
Will you guarantee that in each and any environment a function
returning a pointer uses always the same location as a function
returning int, even as the standard since more than 15 years says the
contrary.
[Insult snipped]

No, that's clearly untrue, no disagreement there, but where is the
undefined behaviour in this program?

#include <stdlib.h>
int main(void) {
free( (char *) malloc(1) );
}

What I said, if you would re-read my message, is that the cast itself
does not invoke undefined behaviour. What does invoke undefined
behaviour is giving an improper declaration of malloc(), but malloc()
is properly declared here. A cast can happen to hide a diagnostic
caused by an improper declaration, but the behaviour is not defined
with or without it.

 

[Ian Collins]

But you says that it is good to write programs for both C anf FORTRAN
or C++ and Cobol?

Where did "programs' come from? The discussion was about headers.

You can't write a single program that uses two highly different
languages. When you needs to write failsave programs instead of
hacking blind around you have to use either C or C++. Mixing up
languages inside the same program ends always up with holes of any
kind.

But they do (often) share headers.

 

[J. J. Farrell]

I would use it as it is defined. Conversion between all kinds of data
pointer to/from pointer to void is done already witrhout cast. When
you does not know what you does you should avoid cats anyway.

Please show us your code to store a void * in a uintptr_t without a
cast.

...

You must learn C before you can use the language, so start learning C.
Until you have learned C you are absolutely unable to something about
the language. You've proven now that you knows nothing about C.

Hmmm ...

Will you guarantee that in each and any environment a function
returning a pointer uses always the same location as a function
returning int, even as the standard since more than 15 years says the
contrary.

How's that relevant?

You knows nothing about C but tells nonsense. There are more traps you
falls in as you can dream of when you does not understund what C is -
and you have not even basic knowlede about C yet.

You need to be on more solid ground than you are to make comments like
these without looking foolish.