Why does comparing a string literal and an integer return a warning of comparison between pointer an-CodePudding

int main()
{
   int c;
   c=getchar();
   if(c=="a")
   {
    printf("fizz");
   }
   else
    printf("buzz");
return 0;
}

Output:

test.c: In function 'main':
test.c:8:8: warning: comparison between pointer and integer
    if(c=="a")

As I understand it, if the string literal "a" was assigned to a variable name a[1] and the comparison was made c==a[0] then the variable name would implicitly decay into *a(0) which is a pointer to the first element in the array. But without a variable name how is the string literal "a" read as a pointer. Does the compiler itself assign a pointer to this string to execute the comparison?

CodePudding user response：

String literals have type "array of char". And like any array, when one is used in an expression it decays (in most cases) to a pointer to its first element.

So the comparison c=="a" is comparing an int on the left and a char * on the right, hence the warning.

CodePudding user response：

String literal is const char array which decays to a pointer.

So your code is almost identical to this one

   const char a[] = "a";
   c=getchar();
   if(c==a)

You need to use character constant

   c=getchar();
   if(c=='a')

or index the string literal

   c=getchar();
   if(c=="a"[0])