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C new malloc question

Why isn't it printed 5

?

void writeValue(int* value) {
    value = malloc(sizeof(int));
    *value = 5;
}


int main(int argc, char * argv) {
    int* value = NULL;
    writeValue(value);
    printf("value = %d\n", *value); // error trying to access 0x00000000
}

and how can I change this to make it work but still use a pointer as an argument for writeValue

?

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3 answers


A pointer ( int *value

) is a value. If you want to keep the above behavior, you need a pointer to pointer.



void writeValue(int** value) {
    *value = malloc(sizeof(int));
    **value = 5;
}


int main(int argc, char * argv) {
    int *value = NULL;
    writeValue(&value); // Address of the pointer value, creates int**
    printf("value = %d\n", *value); // prints 5
}
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There are 2 errors from which I can see: 1) If you want to change the value of your pointer, you need to pass a pointer to a pointer to the value of the int ** function. 2) If you want to print the value of the pointer in your master, you need to override it, * value.



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call malloc before calling writevalue, not inside it (this way you get the added benefit of being able to free it).

Your program doesn't print 5, but it also has a memory leak, losing the address of the allocated block.

The reason, also explained by others, is that the parameter int * value

is a copy int * value

in the main. You can think of this as a local variable to a function. You can only access the location it points to. When you change a value in a function, the other value in main doesn't change.

void writeValue(int* value) {
    *value = 5;
}


int main(int argc, char * argv) {
    int* value = NULL;
    value = malloc(sizeof(int));
    writeValue(value);
    printf("value = %d\n", *value);
    free(value);
}
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