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I am trying to dynamically create an array in a separate function and store some variables inside of that array and I came across the following: how come when I write;

#include <stdio.h> #include <stdlib.h> void foo(int **arr, int N) { *arr = malloc(sizeof(int)); for (int index = 0; index <= 4; index++) { *arr[index] = 1; } } int main() { int *arr; foo(&arr, 5); for (int bar = 0; bar <= 4; bar++) { printf("%d ", arr[bar]); } }

I get this output;

exited, segmentation fault

However, when I run the following;

#include <stdio.h> #include <stdlib.h> void foo(int **arr, int N) { *arr = malloc(sizeof(int)); } int main() { int *arr; foo(&arr, 5); for (int index = 0; index <= 4; index++) { arr[index] = 1; } for (int bar = 0; bar <= 4; bar++) { printf("%d ", arr[bar]); } }

I get this output;

1 1 1 1 1

I am stuck trying to figure out how to fix the first block of code to get it to work, but I seem to have no idea how to do so. Any help would be appreciated in at least understand why a segmentation fault occurs.

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2 Answers 2

Reset to default
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There are a couple of things going on here that are important details.

Evgeny is correct that you need to allocate N things of sizeof int. Otherwise your program is wrong, and will write to invalid memory locations.

The cause of the segment violation took a few minutes to track down. I added printf statements throughout your code so as to be able to see where the segment violation occurred. It showed that the assignment of 1 to the array element was where it was happening.

I then suspected that the binding/order was the issue, so put the parenthesis around the *arr to make it very clear that it was the intent. The segment violation went away after that. I also changed the initialization to index so the assignment results would be easily verifiable.

#include <stdio.h> #include <stdlib.h> void foo(int **arr, int N) { *arr =(int *) malloc(sizeof(int) * N); for (int index = 0; index < N; index++) { printf("before assignment\n"); (*arr)[index] = index; printf("after assignment\n"); } } int main() { int *arr; foo(&arr, 5); for (int bar = 0; bar < 5; bar++) { printf("%d ", arr[bar]); } }

In short, a useful quick and dirty technique is to insert a bunch of printfs to help narrow down where the issue is happening.

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2 Comments

Well that was easier than I thought. Is it because of the way the order of precedence works with pointers '*arr[index]' is treated as '*(arr[index])'?
Take a look at this. stackoverflow.com/questions/51373516/… it talks about array syntax vs pointer syntax.
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malloc requires full memory size, including array size. So the correct code would be

*arr = malloc(sizeof(int) * N);
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I still get a segmentation fault adding the '* N'.

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