Programming Forums > Application Development > C++

conlau · Mar 14th, 2005, 7:51 AM

Hi all,

probably I`m missing out something very basic here, but I`m having problems with using malloc and free under c. I shrinked the affected part of my code to the following.

#include <stdlib.h>

int main(){

char* a;
a = (char *) malloc(25);
a = "test";
printf("%s\n",a);
free(a);
}

It compiles fine, but when I run the code I get a segmentation fault after it prints out "test". I`m running on a linux os btw.

Any help would be greatly appreciated.

Regards

Ooble · Mar 14th, 2005, 12:17 PM

You want:

(Toggle Plain Text)

strcpy(a, "test");
printf("%s\n", *a);

(I think).

Mad_guy · Mar 14th, 2005, 12:33 PM

Urg, this happens to me all the time. DMA is frusterating, and Segmentation violations are lame. But it gets easier.

(Toggle Plain Text)

#include <stdio.h>
#include <malloc.h>

int main(){

char blah[] = "test", *a = (char *) malloc(strlen(blah));
strcpy(a,blah);
printf("%s\n",a);
getchar();
free(a);
}

Also, Ooble, once you've set data at some allocated memory (whethere you're using memset(), or strcpy() or integers or floats, it's all the same) you NEVER refer to it as a pointer. Because there's then actually data at the newly allocated memory.

Go ahead, try to refer to a as a pointer in that printf() statement and it'll STILL die of segmentation violation. I've learned that on my own from too many segementation faults.

I would also like to note about DMA, this: when you're doing DMA, in every type of memory you're allocating like floats or doubles or longs or ints or whatever, you always use a sizeof() statement in the malloc() statement.

In other words:

(Toggle Plain Text)

int *a = (int *)malloc(4); //WRONG
int *a = (int *)malloc(4 * sizeof(int)); //CORRECT
char *a = (char *)malloc(strlen(stuff) * sizeof(char)); //WRONG
char *a = (char *)malloc(strlen(stuff)); //CORRECT

But why? Simple: no matter what platform you're working on, chars are always one byte. ALWAYS. By doing * sizeof(char) you're just wasting system resources. However, on other platforms, ints and floats and longs and such can be a lot more bytes. Normally ints are 4 and I believe longs are 8.

Just thought you should know.

PS: If you ever want to learn to do REALLY cool stuff with Dynamic Memory Allocation, look up Buffer Overflows.

Ooble · Mar 14th, 2005, 12:42 PM

Knew it. I just play with it until it works, usually.

Mad_guy · Mar 14th, 2005, 1:14 PM

Quote:

Originally Posted by Ooble

Knew it. I just play with it until it works, usually.

I normally just try to get it right the first time, because segmentation faults suck anus, and they're not my friend.

Dizzutch · Mar 14th, 2005, 1:20 PM

i started using valgrind and gdb more now i really can't afford to be stuck on segfaults, give 'm a try.

conlau · Mar 21st, 2005, 9:00 AM

Hi all,

many thanks for all your help!

Just noticed that the following modification does not produce a seg fault -

#include <stdlib.h>

int main(){
char* test;
if ( (test = malloc(25)) == NULL){
printf("Error\n");
exit(1);
}
test = "test";

printf ("%s\n",test);
free(test);
}

The only difference here is that the malloc section is checked to see whether a NULL is returned and hence request has failed. I cannot explain why this is not causing any segmentation faults, while the original code is however....

Regards

Mar 14th, 2005, 7:51 AM	#1
conlau Newbie Join Date: Mar 2005 Posts: 3 Rep Power: 0	Segmentation Fault with malloc and free Hi all, probably I`m missing out something very basic here, but I`m having problems with using malloc and free under c. I shrinked the affected part of my code to the following. #include <stdlib.h> int main(){ char* a; a = (char *) malloc(25); a = "test"; printf("%s\n",a); free(a); } It compiles fine, but when I run the code I get a segmentation fault after it prints out "test". I`m running on a linux os btw. Any help would be greatly appreciated. Regards

Mar 14th, 2005, 12:17 PM	#2
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	You want: (Toggle Plain Text) strcpy(a, "test"); printf("%s\n", a); strcpy(a, "test"); printf("%s\n", a); (I think). __________________ Me :: You :: Them

Mar 14th, 2005, 12:33 PM	#3
Mad_guy Hobbyist Programmer Join Date: Oct 2004 Location: Sandstorm, Techno Club Posts: 239 Rep Power: 4	Urg, this happens to me all the time. DMA is frusterating, and Segmentation violations are lame. But it gets easier. (Toggle Plain Text) #include <stdio.h> #include <malloc.h> int main(){ char blah[] = "test", a = (char ) malloc(strlen(blah)); strcpy(a,blah); printf("%s\n",a); getchar(); free(a); } #include <stdio.h> #include <malloc.h> int main(){ char blah[] = "test", a = (char ) malloc(strlen(blah)); strcpy(a,blah); printf("%s\n",a); getchar(); free(a); } Also, Ooble, once you've set data at some allocated memory (whethere you're using memset(), or strcpy() or integers or floats, it's all the same) you NEVER refer to it as a pointer. Because there's then actually data at the newly allocated memory. Go ahead, try to refer to a as a pointer in that printf() statement and it'll STILL die of segmentation violation. I've learned that on my own from too many segementation faults. I would also like to note about DMA, this: when you're doing DMA, in every type of memory you're allocating like floats or doubles or longs or ints or whatever, you always use a sizeof() statement in the malloc() statement. In other words: (Toggle Plain Text) int a = (int )malloc(4); //WRONG int a = (int )malloc(4 * sizeof(int)); //CORRECT char a = (char )malloc(strlen(stuff) * sizeof(char)); //WRONG char a = (char )malloc(strlen(stuff)); //CORRECT int a = (int )malloc(4); //WRONG int a = (int )malloc(4 * sizeof(int)); //CORRECT char a = (char )malloc(strlen(stuff) * sizeof(char)); //WRONG char a = (char )malloc(strlen(stuff)); //CORRECT But why? Simple: no matter what platform you're working on, chars are always one byte. ALWAYS. By doing * sizeof(char) you're just wasting system resources. However, on other platforms, ints and floats and longs and such can be a lot more bytes. Normally ints are 4 and I believe longs are 8. Just thought you should know. PS: If you ever want to learn to do REALLY cool stuff with Dynamic Memory Allocation, look up Buffer Overflows. Last edited by Mad_guy; Mar 14th, 2005 at 1:15 PM.

Mar 14th, 2005, 12:42 PM	#4
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	Knew it. I just play with it until it works, usually. __________________ Me :: You :: Them

Mar 14th, 2005, 1:20 PM	#6
Dizzutch Professional Programmer Join Date: Dec 2004 Location: Worcester, MA Posts: 441 Rep Power: 4	i started using valgrind and gdb more now i really can't afford to be stuck on segfaults, give 'm a try. __________________ naked pictures of you \| PFO F@H stats

Mar 21st, 2005, 9:00 AM	#7
conlau Newbie Join Date: Mar 2005 Posts: 3 Rep Power: 0	Hi all, many thanks for all your help! Just noticed that the following modification does not produce a seg fault - #include <stdlib.h> int main(){ char* test; if ( (test = malloc(25)) == NULL){ printf("Error\n"); exit(1); } test = "test"; printf ("%s\n",test); free(test); } The only difference here is that the malloc section is checked to see whether a NULL is returned and hence request has failed. I cannot explain why this is not causing any segmentation faults, while the original code is however.... Regards

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Programming Forums > Application Development > C++

Segmentation Fault with malloc and free