Question

The code below gives me the following error:

incompatible types when assigning to type’char[10]’ from type char*
lvalue required as decrement operand.

What might cause this?

#include <stdio.h>
#include <string.h>

int main(void)
{
    char *str1="1234";
    char str2[10];

    str2 = str2 + strlen(str1)-1;          //the str pointer is at 3rd position
    char *p = str2+1;                      //since it has to be a valid string, i assigned pointer p to give the null value at the end of the string.
    *p = '\0';

    while(*(str2--) = *(str1++))            //moving the pointer of str down and pointer of str1 up and copy char from str1 to str2

    printf("%s", str2);

    return 0;
}

Answer 1

str2 is an array, not a pointer, so on the left hand side str2 = ... is not a meaningful C expression.

Either str2 must be a pointer, or use an extra variable, like p, to get the expression

char *p = str2 + strlen(str1);

Similarly (str2--) is not meaningful. Use an extra character pointer to str2, and change that.

Think of an array name as a constant pointer. It can’t be changed, but can be used on the right hand side of an = in an expression.