Why is *format a pointer in printf()?

This is a discussion on Why is *format a pointer in printf()? within the C++ Programming forums, part of the General Programming Boards category; Bear with me on this one. I simply don't understand the following description: int printf ( const char * format, ...

  1. #1
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    Why is *format a pointer in printf()?

    Bear with me on this one. I simply don't understand the following description:

    int printf ( const char * format, ... );

    Why is format a pointer here? If I can be pushed in the right direction, I'd greatly appreciate it. Thanks

  2. #2
    Deathray Engineer MacGyver's Avatar
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    Because you're giving it a string. A string in C is a sequence of chars, basically a char array. When you pass an array to a function, in reality you are passing a pointer to the first element of the array.

    When you use a string literal in C, its address is substitued into its place. So when you write something like the following code, the internal behavior is quite different than you might expect:

    Code:
    printf("Hello, world!\n");
    The string "Hello, world!\n" is probably saved in the .data segment of your program or some other area where it resides outside of the .code section. In its place, the address of the start of where the entire string is stored in memory is given to the printf() function. This allow printf() to receive the entire sequence of chars by a pointer.

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    I couldn't have asked for a better explanation. Thanks!

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