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Home/ Questions/Q 7749957
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T11:09:59+00:00 2026-06-01T11:09:59+00:00

I’m a beginner in C and Let’s say I have a code like this:

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I’m a beginner in C and Let’s say I have a code like this:

#include <stdio.h>
void test(char *t)
{
     t++;
    *t = 'e';
}

void main()
{
    char a[] = "anto";
    printf("%c\n",a[1]);
    test(a);
    printf("%c\n",a[1]);
}

This is the sample code, where I am figuring out how pointers work. According to me the statement:

t++;

in the above code will increment the address of array a by 1 char in the calling function test. Fine, now as far I know the * is used to retrieve the object value that the pointer points to.

But weirdly when I change the t++ to

*t++;

I’m getting the same output as before. I’m literally confused with this, the above statement
*t++; should change the contents only know, according to the definition of * operator.

But again this changes the address of t. How come? Where I’m getting the concept wrong?

Thanks in advance

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  1. Editorial Team
    Editorial Team
    2026-06-01T11:10:00+00:00Added an answer on June 1, 2026 at 11:10 am

    The expression *t++ is parsed as *(t++) — the ++ still applies to the pointer, not the contents. The value of t++ is the value of the pointer itself before the increment, while the value of *t++ is what the pointer points to before the increment.

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