Question

I have a confusion related to this program.

#include <stdio.h>

int main(void)
{
        int value = 10;
        char ch = 'A';

        int* ptrValue = &value;
        char* ptrCh = &ch;

        int* ptrValue1 = value;
        char* ptrCh1 = ch;

        printf("Value of ptrValue = %d and value of ptrValue1 = %d\n", *ptrValue, ptrValue1);

        printf("Value of ptrCh = %c and value of ptrCh1 = %c\n", *ptrCh, *ptrCh1);
}

I get two warnings while compiling this program

unipro@ubuguest:/SoftDev/ADSD/Module
1/Unit 1/Rd/C/System$ cc
charPointers.c -o charPointers
charPointers.c: In function
‘main’:
charPointers.c:11:
warning: initialization makes pointer
from integer without a cast
charPointers.c:12: warning:
initialization makes pointer from
integer without a cast

And I know what they mean.

While running the program I get the following error.

unipro@ubuguest:/SoftDev/ADSD/Module
1/Unit 1/Rd/C/System$
./charPointers
Value of ptrValue
= 10 and value of ptrValue1 = 10
Segmentation fault

I know I am getting the error at second printf method.

So my question is if I store a value in pointer, why can’t we de-reference it? Does it behave like a variable?

Thanks.

Answer 1

With the assignment

char* ptrCh1 = ch;

you will set the address to 0x41, not the value. The program will later try to de-reference 0x41 (ie. fetch the data at adress 0x41), which is an illegal address. Hence, the segmentation fault