CODEWITHSUNDEEP
c
Radha Gogia
Radha Gogia

Reputation: 795

why is it legal to declare an inbuilt library function as an identifier name?

This is the declaration of a variable printf:

int printf=90;

This runs without any error , but if I try to print the value of the identifier with the given name printf , then it gives error , so why is it that although the compiler is allowing printf to be used as an identifier but we can't later print its value .

Upvotes: 4

Views: 121

Answers (1)

unwind
unwind

Reputation: 399863

Your question is a bit mixed up.

C has "scope rules", that govern how a name (like printf) is resolved from a given point in the code. It's possible, and legal according to the scope rules, to shadow a name from an outer scope.

Typically printf is extern-declared from <stdio.h>, and its value is replaced by the proper function address from the standard library by the linker.

But if you declare a new variable like that, you are shadowing the name. Basically you get a new variable of a new type, with the same name as an existing one. The outer, original, printf then becomes unreachable from that scope, since C doesn't have a scope resolution operator. In C++, you could use ::printf which always references the global printf. C doesn't have that.

You can work around it by copying the original value of printf before shadowing it:

{
  int (*theprintf)(const char *fmt, ...) = printf;
  int printf = 90;
  theprintf("now printf has the value %d\n", printf);
}

As mentioned in a comment, once you're done with your local variant you can re-declare it as extern to get it back, but you can't do that in the same scope:

{
  extern int printf(const char * fmt, ...);
  printf("hello?\n");
}

Upvotes: 6

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