Thread: Understanding Conversion Specifications!

  1. #1
    Registered User
    Join Date
    Nov 2017
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    34

    Unhappy Understanding Conversion Specifications!

    What's the use of %h , %u , %i when we can simply use %d for all the purposes.

    For example;

    Code:
    #include <stdio.h>
    
    int
    main() {
    printf("%h",78);
    printf("%d",78);
    printf("%i",8909);
    printf("%d",8909);
    printf("%u",65530);
    printf("%d",65530);
    }
    as we can see we can use %d in place of %h,%i,%u , i know that they are used as modifiers for short int(h) , decimal , octal , hexadecimal interger(i) , and unsigned decimal integer(u)

    so what's there purpose? I can't find any benefit or distinction in their usage. Please help!!

  2. #2
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    For the printf family of functions, %i is exactly the same as %d. However, for the scanf family, %i accepts octal input (with a leading 0) and hex input (with a leading 0x), whereas %d is only for decimal. So entering 077 will yield the decimal value 77 with %d but the decimal value 63 with %i.

    %h is again important for scanf but not for printf. The reason it's not important for printf is that, due to printf being a variadic function, any char or short passed to it is converted to an int before being passed.

    %u is for unsigned ints whereas %d is for signed ints, so that's obviously a difference as shown by the following:
    Code:
    #include <stdio.h>
    #include <limits.h>
    
    
    int main() {
        unsigned u = UINT_MAX;
        printf("%u\n", u); // prints 4294967295 (assuming 32-bit ints)
        printf("%d\n", u); // prints -1
        return 0;
    }
    If you want the truth to stand clear before you, never be for or against. - Sent-ts'an

  3. #3
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    Thanks Man, you are a life saver!! Kudos man

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