Macros

Question

# include <stdio.h> # define scanf  "%s GeeksQuiz" int main() {    printf(scanf, scanf);    return 0; }

(A) Compiler Error
(B) %s Geeks Quiz
(C) Geeks Quiz
(D) %s Geeks Quiz Geeks Quiz

Comments

I just modified a little, can u pls tell now how it works?? i mean why GeeksQuiz is not printed now?

 

# include <stdio.h>

# define scanf  "%s GeeksQuiz"

int main()

{

   printf("abcdf",scanf);

   return 0;

}

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First of all, the compiler will throw a warning that too many arguments are supplied to "printf". And I don't think there's any way that "%s GeeksQuiz" will be printed by this program. The only output that'll get printed is "abcdf".

Answer 1

when the program will execute 

printf(scanf, scanf);

both scanf will be replaced by a string "%s GeeksQuiz"

a statement will become printf("%s GeeksQuiz ", "%s GeeksQuiz ");

%s will act as a format specifier, it will print the string "%s GeeksQuiz" and then whatever message u are going to print like in our case, GeeksQuiz will be printed.

the final output will be:%sGeeksQuiz GeeksQuiz  

option D

Comments

why %s is in op in first one or not second one , m not getting ur reason :(

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because %s in first one is telling that there is string supplied as an input which has to be printed wheras %s in second is itself a part of string

in short words  %s in first one is acting as an specifier and %s in second one is a part of string...

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sorry , may be its silly but , if in second one %s is part of string then it should be print in second not first one .?

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yes, %s is printed from second part only...

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ok then due to comma it prints right most 1st then second ... that why its %sgeekquiz geek quiz rt ??

that is last doubt :P

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yes...

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ok thanks for your patience :)

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it is very good thing....'not to accept the answer untill you have grabbed the correct concept'..

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@joshi_nitish @sid1221 The order of execution in printf statement is from right to left ? Can you please explain how the flow of the result getting generated?