Help Problem with Lvalue required

[Maurizio Porcu]

Hello i have this my code but have problem with char=char

# include <conio.h>
# include <stdio.h>
struct Pate {
char Nome[20];
};

main()
{
char lettere[3] = { 'A','P','Z'};
Pate articoli[3]={{"gli"},{"un"},{"una"}};
Pate
preposizioni[10]={{"di"},{"a"},{"da"},{"in"},{"con"},{"su"},{"per"},{"tra"},
{"fra"}};

char text[5000],pat[100], te[1];
int no,count,pos[20],i;
clrscr();
cout<<"\n\nInsert text:";
gets(text);

cout<<"Insert one string:";
// gets(pat);
for (i=0;i<3;i++){
//pat=lettere; //error
pat=articoli[1].Nome; //error
}
}

 

[mlimber]

Hello i have this my code but have problem with char=char

No, you have a problem with char[] = char[]. You'll probably find it
easier to work with std::string.

# include <conio.h>
# include <stdio.h>

conio.h is not standard. stdio.h is for C. C++ should use <cstdio> or

struct Pate {
char Nome[20];
};

How about:

typedef char Nome[20];

or better:

main()

int main()

{
char lettere[3] = { 'A','P','Z'};
Pate articoli[3]={{"gli"},{"un"},{"una"}};
Pate
preposizioni[10]={{"di"},{"a"},{"da"},{"in"},{"con"},{"su"},{"per"},{"tra"},
{"fra"}};

Make these all const.

char text[5000],pat[100], te[1];

An array of one?

int no,count,pos[20],i;
clrscr();
Non-standard.

cout<<"\n\nInsert text:";
gets(text);


cout<<"Insert one string:";
// gets(pat);
for (i=0;i<3;i++){
//pat=lettere; //error
pat=articoli[1].Nome; //error

Use std::string to get this type of convenience. Otherwise, you'll need

}

return 0;

}

Cheers! --M

 

[Victor Bazarov]

Hello i have this my code but have problem with char=char

No, you don't. You have a problem with char[]=char[].

# include <conio.h>
# include <stdio.h>
struct Pate {
char Nome[20];
};

main()

int main()

{
char lettere[3] = { 'A','P','Z'};
Pate articoli[3]={{"gli"},{"un"},{"una"}};
Pate
preposizioni[10]={{"di"},{"a"},{"da"},{"in"},{"con"},{"su"},{"per"},{"tra"},
{"fra"}};

char text[5000],pat[100], te[1];
int no,count,pos[20],i;
clrscr();
cout<<"\n\nInsert text:";
gets(text);

cout<<"Insert one string:";
// gets(pat);
for (i=0;i<3;i++){
//pat=lettere; //error
pat=articoli[1].Nome; //error

Arrays are not assignable in C++

}
}

Don't use character arrays, use 'std::string'.

V

 

[Maurizio Porcu]

give me recommend pelase
i student and have need to solve this problem

Hello i have this my code but have problem with char=char

No, you don't. You have a problem with char[]=char[].

# include <conio.h>
# include <stdio.h>
struct Pate {
char Nome[20];
};

main()

int main()

{
char lettere[3] = { 'A','P','Z'};
Pate articoli[3]={{"gli"},{"un"},{"una"}};
Pate
preposizioni[10]={{"di"},{"a"},{"da"},{"in"},{"con"},{"su"},{"per"},{"tra"},
{"fra"}};

char text[5000],pat[100], te[1];
int no,count,pos[20],i;
clrscr();
cout<<"\n\nInsert text:";
gets(text);

cout<<"Insert one string:";
// gets(pat);
for (i=0;i<3;i++){
//pat=lettere; //error
pat=articoli[1].Nome; //error

Arrays are not assignable in C++

}
}

Don't use character arrays, use 'std::string'.

V

 

[Victor Bazarov]

give me recommend pelase
i student and have need to solve this problem

Don't top-post.

Hello i have this my code but have problem with char=char

No, you don't. You have a problem with char[]=char[].

# include <conio.h>
# include <stdio.h>
struct Pate {
char Nome[20];
};

main()

int main()

{
char lettere[3] = { 'A','P','Z'};
Pate articoli[3]={{"gli"},{"un"},{"una"}};
Pate

preposizioni[10]={{"di"},{"a"},{"da"},{"in"},{"con"},{"su"},{"per"},{"tra"},

{"fra"}};

char text[5000],pat[100], te[1];
int no,count,pos[20],i;
clrscr();
cout<<"\n\nInsert text:";
gets(text);

cout<<"Insert one string:";
// gets(pat);
for (i=0;i<3;i++){
//pat=lettere; //error
pat=articoli[1].Nome; //error

If you just want to transfer all 20 characters from 'articoli[1].Nome'
to 'pat', use 'memcpy':

memcpy(pat, articoli[1].Nome, 20);

V

 

[mlimber]

give me recommend pelase
i student and have need to solve this problem

Read in your C++ book about using std::string. It works something like
this:

#include <iostream>
#include <string>
using namespace std;

int main()
{
cout << "Enter your name: \n";
string name;
getline( cin, name );

if( name == "mlimber" ) // compare two strings
{
name = "chump"; // Assign a new string
}

// A constant string
const string hello = "Ciao, ";

// display the result
cout << hello << name << '.' << endl;

return 0;
}

Cheers! --M

 

[mlimber]

If you just want to transfer all 20 characters from 'articoli[1].Nome'
to 'pat', use 'memcpy':

memcpy(pat, articoli[1].Nome, 20);

Better to use strcpy (or strncpy) with strings.

Cheers! --M

 

[Victor Bazarov]

If you just want to transfer all 20 characters from 'articoli[1].Nome'
to 'pat', use 'memcpy':

memcpy(pat, articoli[1].Nome, 20);

Better to use strcpy (or strncpy) with strings.

No.

'strcpy' can "accidentally" go beyond 20 (if the terminator is missing
in the first 20), and 'strncpy' will append zero chars if it encounters
the zero character before copying 20 chars. So, neither will do what is
needed.

Given

char five[5] = { '1','2','3','4','5' };
char seven[7] = "ab\0cd\0";
char sevenmore[7];

try

strcpy(sevenmore, five); // BOOM! Undefined behaviour

or

strncpy(sevenmore, seven); // what's the result?

V

 

[mlimber]

If you just want to transfer all 20 characters from 'articoli[1].Nome'
to 'pat', use 'memcpy':

memcpy(pat, articoli[1].Nome, 20);

Better to use strcpy (or strncpy) with strings.

No.

'strcpy' can "accidentally" go beyond 20 (if the terminator is missing
in the first 20), and 'strncpy' will append zero chars if it encounters
the zero character before copying 20 chars. So, neither will do what is
needed.

Given

char five[5] = { '1','2','3','4','5' };
char seven[7] = "ab\0cd\0";
char sevenmore[7];

try

strcpy(sevenmore, five); // BOOM! Undefined behaviour

or

strncpy(sevenmore, seven); // what's the result?

V

Good points, although the comp.lang.c FAQ advocates using strcpy (8.3
and the second paragraph of 13.2). In any case, these are just more
reasons to prefer std::string.

Cheers! --M

 

[Aleksey Loginov]

If you just want to transfer all 20 characters from 'articoli[1].Nome'
to 'pat', use 'memcpy':

memcpy(pat, articoli[1].Nome, 20);

or even std::copy ( articoli[1].Nome, articoli[1].Nome+20, pat );

 

[Maurizio Porcu]

i has used the memcpy(pat, articoli[0].Nome, 3);
but if use pat and sent this in a function not go
if i use pat with gets(pat) the function go well

 

[mlimber]

i has used the memcpy(pat, articoli[0].Nome, 3);
but if use pat and sent this in a function not go
if i use pat with gets(pat) the function go well

You're posting to the C++ forum, yet you're using C-style techniques.
You should either post to comp.lang.c or follow our previous suggestion
to use std::string instead of character arrays.

Cheers! --M

 

[Maurizio Porcu]

yes but i use borland c++ and i have to write in c++
and i have problem with this source

i has used the memcpy(pat, articoli[0].Nome, 3);
but if use pat and sent this in a function not go
if i use pat with gets(pat) the function go well

You're posting to the C++ forum, yet you're using C-style techniques.
You should either post to comp.lang.c or follow our previous suggestion
to use std::string instead of character arrays.

Cheers! --M

 

[mlimber]

yes but i use borland c++ and i have to write in c++
and i have problem with this source

Sure, but C++ is a superset of C (mostly), which means that most C
programs will build with under C++. In other words, you can use a C++
compiler to write C programs, and, in fact, that seems to be what you
are doing. However, the C approach to solving problems often varies
widely from the C++ apprach because the latter relies on the C++
standard library (including std::string) and OO abstraction, which are
not available in C. Since our advice here will be in terms of those
standard C++ facilities, I'd suggest that if you prefer C techniques
(even if you're using a C++ compiler), you'll get more helpful advice
in comp.lang.c.

Cheers! --M

PS, It's considered bad manners to put your replies at the top of
previous posts as you did with this one.