Why does this C code cause segmentation error? Trying to understand pointers

Question

Trying to understand what exactly happens in the memory during the execution of this code and why it causes a segmentation error.

#include <stdio.h>

int main() {
    int* ptr;
    *ptr = 5;
    printf("%d", *ptr);
    return 0;
}

`*ptr = 5` attempts to write `5` to the address held in `p`. Questions: what _is_ this address? Do you have permission to write to it? — ForceBru, May 03 '19 at 16:34

score 2 · Accepted Answer · answered May 03 '19 at 16:33

All variables must first be initialized to a valid value before you can use them in an expression like *p. For pointers, this means assigning an address to point to. You can do this either by taking the address of another variable or by dynamically allocating memory with malloc().

When you try to dereference an uninitialized pointer with *p, it will use whatever memory address happens to be stored in p at that time. Most likely this address is invalid for you to use and so you get a segfault.

score 0 · Answer 2 · edited May 03 '19 at 16:34

0

First allocate memory to the pointer variable

ptr = (int* ) malloc(sizeof(int))

edited May 03 '19 at 16:34

ForceBru

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answered May 03 '19 at 16:34

Mani2019

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The OP was asking why a seg fault was happening. This will fix the seg fault, but doesn't really explain what's going on in the original bit of code. – Michael Albers May 03 '19 at 16:35
Please change it as ptr = (int*) malloc(sizeof(int)); – Mani2019 May 03 '19 at 16:36
@Mani You can edit your answer yourself. – Code-Apprentice May 03 '19 at 16:37
1

@Mani Why do you want `(int*)`? It's completely unnecessary. Also, you could change `sizeof(int)` to `sizeof(*ptr)`. – klutt May 03 '19 at 16:45

Why does this C code cause segmentation error? Trying to understand pointers

2 Answers2