[Question] C Program crashes after malloc

[d3rrial]

Hey Guys, I have a problem: My Program always crashes when I change values inside a variable that I allocated with malloc. I can't even free the program because it will just crash.

I also I can't find anything wrong with the code, so could someone please explain to me why my Program always crashes? This is quite important, tonight is the deadline

Thanks alot!

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

#define Y 30
int nrreturn();

int main(){
	char** feld;
	int i,j, z;
	feld=(char**)malloc(sizeof(char*)*nrreturn());
	for(i=0;i<Y;i++){
		feld[i]=(char*)malloc(sizeof(char)*nrreturn());
	}
	for(j=0;j<Y;j++){
		for(i=0;i<Y;i++){
			feld[j][i]=0;
		}
	}
	
	for(j=0;j<nrreturn();j++){
		for(i=0;i<Y;i++){
			printf("%d",feld[j][i]);
		}
		while(z<1000){
			z++;
		}
		z=0;
		free(feld[j]);
		printf("\n");
	}
	free(feld);
	return 0;
}

int nrreturn(){
	return 5;
}

The Console Output:

X:\perm\perm2>main
000000000000000000000000000000

Then after that there's an error message, that the program isn't reacting anymore.

Thanks alot again!

Edit 2: This is not the real program, this is more or less a debugging program I used here that is supposed to show the effects of my real program. (Which for example uses a value gathered from a function to determine how many rows the array is supposed to have)

[macpain78]

for(j=0;j<Y;j++){

You allocated nrreturn not Y.

So when you try to acess feld[5][0] it segfault.
The correct statement would be:
for(j=0;j< nrreturn() ;j++){

Which, you wrote 3 lines later

EDIT: feld[5] not 6 sry

[d3rrial]

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

#define X 5
#define Y 20

int main(){
	int **feld;
	int i, a;
	
	feld = (int**)malloc(X*sizeof(int*));
	
	for(i=0;i<5;i++){
		feld[i]=(int*)malloc(Y*sizeof(int));
	}
	
	feld[0]=(int*)"Hallo";
	feld[1]=(int*)"Welt!";
	feld[3]=(int*)"\n";
	printf("%s %s%s",feld[0],feld[1],feld[3]);
	for(i=0;i<X;i++){
		free(feld[i]);
	}
	free(feld);
	return EXIT_SUCCESS;
}

That does the very same error. It just crashes in the middle of freeing the allocated memory!

Also I just used int** this time to demonstrate that this is not a problem of char**

[_Mike]

First, why are you casting the return from malloc? And if your answer is "because I have to" then this isn't C.
What compiler are you using?

Secondly,

feld[0]=(int*)"Hallo";
feld[1]=(int*)"Welt!";
feld[3]=(int*)"\n";

You are overwriting your pointers to the allocated memory with pointers to (in the context of free() ) garbage. And again, why the casts? If you have to cast something to get around a compile error then that's often a big hint that you are doing something terribly wrong.
Strings should be pointer to char. And to be strict, string literals should be pointer to const char.

Thirdly, using the = operator with a string does not copy the string itself. It copies the pointer to the string. Look up strcpy() or in your case preferably strncpy() since you are using fixed length buffers.

And from the printf in your first post; %d expects an int-sized argument, not a char. See printf format string - Wikipedia, the free encyclopedia the length field specifically.

[Arlyh]

I hope this helps you.

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

//#define Y 30  //This is unused
int nrreturn();

int main(int argc, char **argv) {
	char **feld;
	int i, j, z = 0;  //it is dangerous to use an uninitialized variable, z could be set to a very high value, causing no loop or a very long loop based on sign.
	feld = (char **)malloc(sizeof(char *) * nrreturn());
	//for(i = 0; i < Y; i++) {	//Y = 30, but feld only has enough memory allocated for nrreturn() elements
	for(i = 0; i < nrreturn(); i++) {	
		feld[i] = (char *)malloc(sizeof(char) * nrreturn());
	}

	//This has the same problem, you've only allocated 5 elements of 5 char's each, but you're assigning 30 elements of 30 char's.
	/*for(j = 0; j < Y; j++) {
		for(i = 0; i < Y; i++) {*/
	for(j = 0; j < nrreturn(); j++) {
		for(i = 0; i < nrreturn(); i++) {
			feld[j][i] = 0;
		}
	}
	
	for(j = 0; j < nrreturn(); j++) {
		//for(i = 0; i < Y; i++) {	//same
		for(i = 0; i < nrreturn(); i++) {	
			printf("%d",feld[j][i]);
		}
		//why are we doing this?
		while(z<1000) {  
			z++;
		}
		z = 0;
		free(feld[j]);
		printf("\n");
	}
	free(feld);
	return 0;
}

int nrreturn() {
	return 5;
}

printf("%d", some_char) has worked forever, and I believe that printf() properly promotes the char to an int when used in this way. If you believe otherwise, you can just cast the char to an int.

Assuming you're using Visual Studio, go to the properties page for your project, navigate to C/C++ -> Advanced, locate the Compile As property and change it to Compile as C. Otherwise just call this code C++, so no one gets confused.

[_Mike]

printf("%d", some_char) has worked forever, and I believe that printf() properly promotes the char to an int when used in this way. If you believe otherwise, you can just cast the char to an int.

Yes integer promotion would normally apply for function arguments, except the standard says that if the function prototype ends with ... then the promotion behavior is undefined. But on all the compilers I've tried it works anyways. Normally I wouldn't care but when someone is trying to learn the language they should be aware of the possible limitations. Once they understand the rules they are free to start breaking them

[Arlyh]

Yes integer promotion would normally apply for function arguments, except the standard says that if the function prototype ends with ... then the promotion behavior is undefined. But on all the compilers I've tried it works anyways. Normally I wouldn't care but when someone is trying to learn the language they should be aware of the possible limitations. Once they understand the rules they are free to start breaking them

I was actually looking for that before I replied but couldn't find it. I'm sure you're right regardless, as I was told the same thing a number of years ago. Thanks for explaining the issue

Additionally, OP, if you could explain what you are trying to accomplish with the while(z < 1000) loop, perhaps we could help you find a better way to do it. I assume you're using this as a delay, but it's first a busy loop, so you're using system resources while it runs, and secondly it's too short to really accomplish much of a delay. All major operating systems have a sleep(uint ms) function, check windows.h for windows, and unistd.h (I believe) for linux.

[d3rrial]

First, why are you casting the return from malloc? And if your answer is "because I have to" then this isn't C.
What compiler are you using?

I am using the Visual Studio compiler

Secondly,

feld[0]=(int*)"Hallo";
feld[1]=(int*)"Welt!";
feld[3]=(int*)"\n";

You are overwriting your pointers to the allocated memory with pointers to (in the context of free() ) garbage. And again, why the casts? If you have to cast something to get around a compile error then that's often a big hint that you are doing something terribly wrong.
Strings should be pointer to char. And to be strict, string literals should be pointer to const char.

Thirdly, using the = operator with a string does not copy the string itself. It copies the pointer to the string. Look up strcpy() or in your case preferably strncpy() since you are using fixed length buffers.

I was casting because I didn't expect an Integer array being able to handly char array values correctly, I could have done it without it but I guess I did it as a "safety measure"
Also: I didn't know that the = operator overwrites the pointer. Thanks a lot, that pretty much solved my problem! I know strcopy(), thanks.
Edith: And I used casts, because I just learned it that way.

And from the printf in your first post; %d expects an int-sized argument, not a char. See printf format string - Wikipedia, the free encyclopedia the length field specifically.

I know that printf would expect an Integer argument, but if I had used char, I would have "outputted" (sorry I dont know the correct translation) the actual char stored in there but not the numeric value.