How does c compare character variable against string?

Question

The following code is completely ok in C but not in C++. In following code if statement is always false. How C compares character variable against string?

Answer 1

I wouldn't say the code is okay in either language.

'a' is a single character. It is actually a small integer, having as its value the value of the given character in the machine's character set (almost invariably ASCII). So 'a' has the value 97, as you can see by running

char c = 'a';
printf("%d\n", c);

"a", on the other hand, is a string. It is an array of characters, terminated by a null character. In C, arrays are almost always referred to by pointers to their first element, so in this case the string constant "a" acts like a pointer to an array of two characters, 'a' and the terminating '\0'. You could see that by running

char *str = "a";
printf("%d %d\n", str[0], str[1]);

This will print

97 0

Now, we don't know where in memory the compiler will choose to put our string, so we don't know what the value of the pointer will be, but it's safe to say that it will never be equal to 97. So the comparison if(ch=="a") will always be false.

When you need to compare a character and a string, you have two choices. You can compare the character to the first character of the string:

if(c == str[0])
     printf("they are equal\n");
else printf("confusion\n");

Or you can construct a string from the character, and compare that. In C, that might look like this:

char tmpstr[2];
tmpstr[0] = c;
tmpstr[1] = '\0';

if(strcmp(tmpstr, str) == 0)
     printf("they are equal\n");
else printf("confusion\n");

That's the answer for C. There's a different, more powerful string type in C++, so things would be different in that language.