printing an array of strings using pointers only in C [duplicate]

Array designators are non-modifiable lvalues. You may not change them. What you need is the following

static const char *Props[] = {"Cp", "Cv", "Mu", "H"};

for ( size_t i = 0; i < sizeof( Props ) / sizeof( *Props ); i++ )
{
    puts( Props[i] );
}

Another approach is to add a sentinel value to the array. For example

static const char *Props[] = {"Cp", "Cv", "Mu", "H", NULL };

for ( const char **s = Props; *s; s++ )
{
    puts( *s );
}

It doesn't work work the same reason as int array [] = {1,2}; array++; doesn't work.

You can't apply ++ on an array type. You would need a pointer to the first item of the array instead. Thus one solution would have been to do this:

const char** ptr = &Props[0]; 
while(*ptr) printf("%s\n", *ptr++);

But that's pretty horrible code and also contains another bug, namely the lack of an end condition in the array. For this solution to work, the array should have been declared as {"Cp", "Cv", "Mu", "H", NULL};.

Don't do weird things like this just because you can though. Just use a for loop with an integer iterator, as demonstrated in another answer.

Only for the purposes of satisfying the requirements of this odd question:

#include <stdio.h>

static void
print_props(void)
{
    static const char *Props[] = {"Cp", "Cv", "Mu", "H"};
    const char **b = Props;
    const char **e = Props + sizeof(Props)/sizeof(Props[0]);

    do {
        puts(*(b++));
    } while (b != e);
    return;
}

int
main(void)
{
    print_props();
    return 0;
}

However, just use a for loop.