copy char*

[Gary Wessle]

string a = "abcdef";
aa[80] = a.c_str();

bb[40] = how can I get def from aa in here?

thanks

 

[ondra.holub]

Gary Wessle napsal:

string a = "abcdef";
aa[80] = a.c_str();

bb[40] = how can I get def from aa in here?

thanks

a.c_str() + 4

But you should know, that result of c_str() method is valid only till
the first change in `a'. So you should make copy of i if you need
C-string and cannot guarantee, that no change in `a' will occure.

 

[SirMike]

Dnia 15 Nov 2006 22:43:39 +1100, Gary Wessle napisa³(a):

string a = "abcdef";
aa[80] = a.c_str();

bb[40] = how can I get def from aa in here?

use strncpy

 

[Default User]

string a = "abcdef";
aa[80] = a.c_str();

bb[40] = how can I get def from aa in here?

What are aa and bb?



Brian

 

[Gary Wessle]

Dnia 15 Nov 2006 22:43:39 +1100, Gary Wessle napisał(a):

string a = "abcdef";
aa[80] = a.c_str();

bb[40] = how can I get def from aa in here?

use strncpy

void func1(const char* p){
char frst[40];
char last[40];
strncpy(frst, p, 3); //first 3 ????
strncpy(last, p, 3); //last 3 ?????
}

string a = "abcdef";
func1(a.c_str());

 

[Gary Wessle]

I want frst to have "abc" and last to have "def".

 

[Jim Langston]

Dnia 15 Nov 2006 22:43:39 +1100, Gary Wessle napisal(a):

string a = "abcdef";
aa[80] = a.c_str();

bb[40] = how can I get def from aa in here?

use strncpy

void func1(const char* p){
char frst[40];
char last[40];
strncpy(frst, p, 3); //first 3 ????
strncpy(last, p, 3); //last 3 ?????
}

string a = "abcdef";
func1(a.c_str());

void func1( const char* p )
{
char first[40];
char last[40];
strncpy( first, p, 3 );
strncpy( last, p + 3, 3 );
}

Ugly code. Prefer std::string instead.

void func1( const std::string Value )
{
std::string first = Value.substr( 0, 3 );
std::string last = Value.substr( 4, 3 );
}

 

[BobR]

Gary Wessle wrote in message ...

use strncpy

void func1(const char* p){
char frst[40];
char last[40];
strncpy(frst, p, 3); //first 3 ????
strncpy(last, p, 3); //last 3 ?????
}

string a = "abcdef";
func1(a.c_str());

void func1( char const *p, std:stream &out ){
char frst[ 40 ];
char last[ 40 ];
strncpy( frst, p, 3 ); //first 3 ????
strncpy( last, p+3, 3 ); //last 3 ?????
out << "frst="<<frst<<" last="<<last<<std::endl;
return;
}

{
std::string a = "abcdef";
func1( a.c_str(), std::cout );
}
// out: frst=abc last=def

 

[Default User]

void func1( char const *p, std:stream &out ){
char frst[ 40 ];
char last[ 40 ];
strncpy( frst, p, 3 ); //first 3 ????
strncpy( last, p+3, 3 ); //last 3 ?????
out << "frst="<<frst<<" last="<<last<<std::endl;
return;
}

{
std::string a = "abcdef";
func1( a.c_str(), std::cout );
}
// out: frst=abc last=def

Did you actually try this? If you got the purported results, then it
was by sheer bad luck. You didn't terminate the string, and strncpy
doesn't do it for you. Outputting the result was undefined behavior.




Brian

 

[Gary Wessle]

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

class myType {
protected:
char _frst[40];
char _second[40];
char _pair[80];
public:
const char* frst() const;
const char* second() const;
myType(const char* pair);
};

myType::myType(const char* pair)
{
strncpy(_frst, pair, 3);
strncpy(_second, pair+4, 3);
}
const char* myType::frst()const {return _frst;}
const char* myType::second()const {return _second;}

int main(){
string s = "abc-def";
myType mt(s.c_str());
string tmp = mt.frst();
string tmp2 = mt.second();
cout << tmp << " " << tmp2 << endl;
}

//output
abcMh Q%G.ANoN?N=NoN?N=NoN?N=NoN?N=%@`%G.ANoN?N=%@M$B'\(B$B%G.ANoN?N=(B$B%@(B def

 

[Old Wolf]

string a = "abcdef";
aa[80] = a.c_str();

This is an error: you can't assign to arrays

bb[40] = how can I get def from aa in here?

Assuming you have managed to get "a" into "aa" successfully:
memcpy(bb, aa + 3, 3);

 

[Gary Wessle]

I need it to ouput
abc def

what must be done to
strncpy(_frst, pair, 3);
strncpy(_second, pair+4, 3);
in the c'tor?

 

[BobR]

BobR wrote:


Did you actually try this?
Yes.

If you got the purported results, then it
was by sheer bad luck. You didn't terminate the string, and strncpy
doesn't do it for you. Outputting the result was undefined behavior.

Of course you are correct. My bad. (gads, I hate that phrase!!).
Thanks for the 'heads-up'.

<?>
It must have been that the char arrays were initialized to zeros by my
compiler.
<checking....>
Nope!!! Just happened to be a zero at 4th position in both arrays.
[....even on re-compiles. ....and adding another array before those.]
[ I can't seem to break it! ]
</?>

Murphy's "bad luck" clause strikes again!!

Thanks again for the correction.

 

[BobR]

BobR wrote:

Outputting the result was undefined behavior.

[ ref: the missing terminator. ]

HEY! How come why for you yell at me but not Jim?
<G>

 

[Gary Wessle]

in order to get the clean output "abd def" I did some changes to the
code but now getting Seg Fault.

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

class myType {
protected:
char _frst[40];
char _second[40];
char _pair[80];
public:
const char* frst() const;
const char* second() const;
myType(const char* pair);
};

myType::myType(const char* pair)
{
strncpy(_frst, pair, 3);
strncpy(_second, pair+4, 3);
}
const char* myType::frst()const {
char* r;
strncpy(r,_frst,3);
return r;}
const char* myType::second()const {
char* r;
strncpy(r,_second,3);
return r;
}

int main(){
string s = "abc-def";
myType mt(s.c_str());
cout << mt.frst() << " " << mt.second() << endl;
}

 

[BobR]

Gary Wessle wrote in message ...

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

class myType { protected:
char _frst[40];
char _second[40];
char _pair[80];
public:
const char* frst() const;
const char* second() const;
myType(const char* pair);
};

myType::myType(const char* pair){
strncpy(_frst, pair, 3);

// Like Default User pointed out, I forgot to terminate the array:
_frst[4] = '\0';

strncpy(_second, pair+4, 3);

_second[4] = '\0';

// or:
// frst[4]= 0;
// second[4]= 0;

// also, BAD nameing.
// Put the underscore on the end (not the begin) of the names
// if you MUST use them.

}
const char* myType::frst()const {return _frst;}
const char* myType::second()const {return _second;}

int main(){
string s = "abc-def";

// You changed the spec on us!
// I'll let you figure out the 'fix' as punishment.

myType mt(s.c_str());
string tmp = mt.frst();
string tmp2 = mt.second();
cout << tmp << " " << tmp2 << endl;
}

//output
abcMh

Q%G.ANoN?N=NoN?N=NoN?N=NoN?N=%@`%G.ANoN?N=%@M$B'\(B$
B%G.ANoN?N=(B$B%@(B def

You were luckier than I was!! <G>

 

[BobR]

Gary Wessle wrote in message ...

in order to get the clean output "abd def" I did some changes to the
code but now getting Seg Fault.

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

class myType {
protected:
char _frst[40];
char _second[40];
char _pair[80];
public:
const char* frst() const;
const char* second() const;
myType(const char* pair);
};

myType::myType(const char* pair){
strncpy(_frst, pair, 3); _frst[ 4 ] = 0;
strncpy(_second, pair+4, 3); _second[ 4 ] = 0;
}

const char* myType::frst()const {
char* r;

OK, when working with raw arrays, you always need to ask yourself, "Is there
enough room in there?".
You declared 'r' as a pointer to type 'char', but, where is the storage for
the 'strncpy' to copy to? You need to make some storage, and then put the
address of that into 'r'.
You are getting the 'Seg Fault' because you did not point 'r' to anything, or
initialize it. It's pointing to something that is not yours to change.

Now before you get the idea to do:

char *r( new char[ 40 ] );

....remember that now you are putting the life of that pointed to storage in
the callers hands, it's up to them to see that it is eventually
'delete[]'ed!!
Hint: 'cout' don't know how to fo that!
The easy fix is to dump the 'strncpy' and 'char*' here, and just do:

return _frst;


// > strncpy(r,_frst,3);
// > return r;
}

const char* myType::second()const {
char* r;

Same here.

strncpy(r,_second,3);
return r;
}

int main(){
string s = "abc-def";
myType mt(s.c_str());
cout << mt.frst() << " " << mt.second() << endl;
}

Alf P. Steinbach's "Pointers" document:
http://home.no.net/dubjai/win32cpptut/special/pointers/ch_01.pdf

 

[Jim Langston]

BobR wrote:

Outputting the result was undefined behavior.

[ ref: the missing terminator. ]

HEY! How come why for you yell at me but not Jim?
<G>

Because I didn't try to output the result. I just gave the op what he asked
for, abc in the first, cdf in the second. He didn't ask to terminate them,
nor even why he was doing it. Now, if he actually wants to output them, I
would do it a different way. I suggested he use std::string anyway.

 

[BobR]

Jim Langston wrote in message ...

Default User wrote in message ...

Outputting the result was undefined behavior.

[ ref: the missing terminator. ]

HEY! How come why for you yell at me but not Jim?
<G>

Because I didn't try to output the result.

I suggested he use std::string anyway.

Must be some 'learning' thing, std::string would be too easy.

 

[Default User]

string a = "abcdef";
aa[80] = a.c_str();

This is an error: you can't assign to arrays

But you can to an array element, which is why his missing declarations
for things like aa made the problem insoluable.

If aa were declared something like:

const char *aa[81];

Then you could assign the results from c_str() to that element. Now,
that's apparently not what he was going for.




Brian