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Home/ Questions/Q 8603061
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T02:14:08+00:00 2026-06-12T02:14:08+00:00

#include<stdio.h> int swap(int *a,int *b); int main() { int a=10,b=20; swap(&a++,&b++); printf(a=%d\nb=%d,a,b); return 0;

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#include<stdio.h>
int swap(int *a,int *b);
int main()
{
    int a=10,b=20;
    swap(&a++,&b++);
    printf("a=%d\nb=%d",a,b);
    return 0;
}

int swap(int *a,int *b)
{
    int temp;
    temp=*a;
    *a=*b;
    *b=temp;
}

Why does this function give the error “invalid lvalue in unary ‘&'”?
Normal swap(&a,&b) works fine but swap(&a++,&b++) as well as swap(&(a++),&(b++)) give errors. What’s the reason behind this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T02:14:09+00:00Added an answer on June 12, 2026 at 2:14 am

    The post-increment operator returns a temporary version of the previous value contained in the variable on which the post-increment operation was performed. This temporary value is not a l-value, or “named” memory location, therefore you can’t take the address of that temporary using the unary address-of operator.

    For instance, on certain architectures like x86, etc., a temporary value generated from the post-increment operator on a simple POD-type like a int, long etc. will be temporarily held in a CPU register, not an actual memory location. In these instances you simply can’t take the “address” of a CPU register.

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