Thread: How does the function strcat use?

Code:

 
    char *p1;
    char *p2;
    p1 = "tttttttttt";
    p2 = "sssssssss";
    
    char *p3;
    p3 = strcat(p1,p2);
    
    printf("%s\n",*p3);

result:
error !

why ?
How to solve this problem?

Originally Posted by zcrself

Code:

 
    p3 = strcat(p1,p2);

result:
error !

why ?
How to solve this problem?

What you've attempted here is actually quite reasonable -- it's just not the way strcat() actually works. If strcat() allocated a NEW string, concatenated p1 and p2 together as this new string, and returned it, your code would be fine.

However what strcat() actually does is append p2 directly to the end of p1, thereby ALTERING p1. Problem is, you can't alter p1 for the reasons tabstop mentioned: 1) the string it read-only and 2) it's not big enough even if you COULD write to it.

Originally Posted by tabstop

p1 points to static memory which (1) can't be changed and (2) even if you could, would not have enough room to store more letters. Presumably you either want to (a) use arrays or (b) use malloc et al.

how to solve?
[code]
char *p = malloc(100);

I was going through this thread and did this code:

Code:

#include <stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
	char *p1,*p2;
	p1=malloc(50);
	p2=malloc(50);
	p1="my name is ben10";
	p2="i like cboard and c";
	puts(p1);
	puts(p2);
	strcpy(p1,p2);
	printf("%s	%s",p1,p2);
}

I'm completely frustrated and in shock that why the above code is giving segfault even if I've malloced p1 and p2. Please help here as this will be helpful for the OP as well as me too. I think I'm missing something very minor.

EDIT: If I also not initialize p1, then this program works fine. Is mallocing p1 doesn't make it writable?

Originally Posted by BEN10

I'm completely frustrated and in shock that why the above code is giving segfault even if I've malloced p1 and p2.

Take a look at this snippet:

Code:

p1=malloc(50);
p2=malloc(50);
p1="my name is ben10";
p2="i like cboard and c";

From the first two lines, we see that p1 and p2 point to dynamically allocated memory (or to null pointers if malloc fails to allocate). In the next two lines, we see that p1 and p2 now point to the first character of string literals. Clearly, this is a problem: you have a memory leak, and your intention to avoid an attempt to change a string literal with strcat() is thwarted. A correct solution would be to use strcpy() to copy over the string literals.

Originally Posted by BEN10

I was going through this thread and did this code:

Code:

#include <stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
	char *p1,*p2;
	p1=malloc(50);
	p2=malloc(50);
	p1="my name is ben10";
	p2="i like cboard and c";
	puts(p1);
	puts(p2);
	strcpy(p1,p2);
	printf("%s	%s",p1,p2);
}

I'm completely frustrated and in shock that why the above code is giving segfault even if I've malloced p1 and p2. Please help here as this will be helpful for the OP as well as me too. I think I'm missing something very minor.

EDIT: If I also not initialize p1, then this program works fine. Is mallocing p1 doesn't make it writable?

occur error!
why ?

Originally Posted by BEN10

If I also not initialize p1, then this program works fine. Is mallocing p1 doesn't make it writable?

As I noted, it is the later assignment of "my name is ben10" to p1 that is the problem. I also notice that you did not actually demonstrate strcat(), but rather strcpy(). A more correct, though not very useful example would be:

Code:

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

int main(void)
{
    const char part1[] = "my name is ben10";
    const char part2[] = ", i like cboard and c";
    char *str = malloc(sizeof(part1) + sizeof(part2) - 1);
    if (str)
    {
        strcpy(str, part1);
        puts(str);
        strcat(str, part2);
        puts(str);
        free(str);
    }
    return 0;
}

I tried you example and it worked properly but I still dont understand why this is not working?

Code:

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

int main(void)
{
    const char part1[] = "my name is ben10";
    const char part2[] = ", i like cboard and c";
    char *str = malloc(sizeof(part1) + sizeof(part2) - 1);
	str="why is this wrong";//adding this line results in segf
    if (str)
    {
        strcpy(str, part1);
        puts(str);
        strcat(str, part2);
        puts(str);
        free(str);
    }
    return 0;
}

Please explain me in layman jargon. My main problem is that after malloc'ing str why is that it isn't behaving like an ordinary string.
EDIT: The above code doesn't work but this one works fine. If str after mallocing is still read only then why this works.

Code:

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

int main(void)
{
    const char part1[] = "my name is ben10";
    const char part2[] = ", i like cboard and c";
    char *str = malloc(sizeof(part1) + sizeof(part2) - 1);
    if (str)
    {
        strcpy(str, part1);
        puts(str);
        strcat(str, part2);
        puts(str);
		*str='g'; //changing str
		puts(str);
        free(str);
    }
    return 0;
}

Originally Posted by BEN10

My main problem is that after malloc'ing str why is that it isn't behaving like an ordinary string.

You basically discarded the effects of the malloc. The pointer no longer points to what was returned by malloc; it points to the first character of the string literal. It is as if the malloc was simply removed (but it has not, thus the memory allocated by malloc remains allocated, though unusable).

Put it another way. It is like writing this code:

Code:

int n = 10;
n = 123;

then asking why the value of n is no longer 10 after the second line.

Originally Posted by laserlight

You basically discarded the effects of the malloc. The pointer no longer points to what was returned by malloc; it points to the first character of the string literal. It is as if the malloc was simply removed (but it has not, thus the memory allocated by malloc remains allocated, though unusable).

Put it another way. It is like writing this code:

Code:

int n = 10;
n = 123;

then asking why the value of n is no longer 10 after the second line.

Thanks laserlight. Thanks, Thanks, Thanks......................Now I've understood it. Please give me some part of your sharp brain, fools like me need it.