Memset and characters - c

Question

I aim to copy source string to dest string. If i compile the following program:
#include <stdio.h>
int main(void) {
char dest[6];
char source[6];
strcpy(dest,source);
while (*dest) { printf("%c",*dest++); }
while (*source) {printf("%c",*source++); }
return 0;
}
I get a runtime error. I suspect it is because strcpy copies from source to destination till it encounters \0. It did not,however,encounter the null character and kept on copying from the buffer till the runtime error occurred. To solve this problem, i modified the code as follows:
#include <stdio.h>
int main(void) {
char dest[6];
char source[6];
memset(dest, '\0', 6*sizeof(dest)); //trying to set dest to '/0'
strcpy(dest,source);
while (*dest) { printf("%c",*dest++); }
while (*source) {printf("%c",*source++); }
return 0;
}
i get the following errors:
prog.c:11:38: error: lvalue required as increment operand
while (*dest) { printf("%c",*dest++); }
^
and
prog.c:11:38: error: lvalue required as increment operand
while (*dest) { printf("%c",*source++); }
^
Why does this happen?

Answer 1

For starters it is the source array that shall be zero terminated if you are going to copy it in another character arrays using the standard C function strcpy. So instead of this statement
memset(dest, '\0', 6*sizeof(dest));
you should at least write
memset(source, '\0', 6*sizeof(source));
^^^^^^ ^^^^^^^
However even this statement is wrong because it overwrites the memory allocated for the array. sizeof( source ) is already equal to 6 bytes as it is followed from the array declaration
char source[6];
Thus you have to write
memset(source, '\0', sizeof(source));
^^^^^^^^^^^^^
In fact there was enough to write either like
char source[6] = { '\0' };
or like
char source[6] = "";
or like
char source[6];
source[0] = '\0';
Arrays are non-modifiable lvalues. Thus you may not write for example the following way
while (*dest) { printf("%c",*dest++); }
Instead of this statement you could write
for ( char *p = dest; *p; ++p ) { printf("%c", *p); }
Take into account that nothing will be outputted because the array contains an empty string. You could initialize the source array with some non-empty string literal.

Answer 2

the following code cleanly compiles, and performs the desired operation.
The differences between the posted code and this are commented.
#include <stdio.h> // printf()
#include <string.h> // strcpy()
int main(void)
{
char dest[6]; // declared, containing garbage
char source[6] = "12345"; // declared, containing the string "12345\0"
strcpy(dest,source);
// now both arrays contain the string "12345\0"
// best to use a 'for()' statement for indexing through an array
for( size_t i=0; dest[i]; i++ ) { printf("%c", dest[i]); }
printf( "\n" ); // output the buffered data to the terminal
for( size_t i=0; source[i]; i++ ) { printf("%c", source[i]);}
printf( "\n" ); // output the buffered data to the terminal
// note, the following lines contain a precedence problem in
// the increment expressions and
// the address of an array declaration cannot be incremented
//while (*dest) { printf("%c",*dest++); }
//while (*source) {printf("%c",*source++); }
//return 0;// with modern C compilers,
// this line is not necessary in a 'main()' function
// when returning 0
} // end function: main

Answer 3

strcpy is not a safe function, prefer using strncpy.
The error is due to the fact that you try to increment the array, which is an rvalue (i.e. a constant, you can not put it to the left side of a sign =).
The common approach to iterate over an array is to use a pointer like so:
char *p = dest;
while (*p) { printf("%c",*p++); }