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What will be the output of the program? #include int main() { const char *s = ""; char str = "Hello"; s = str; while(*s) printf("%c", *s++); return 0; }

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1. 

What will be the output of the program?

#include<stdio.h>

int main()
{
const char *s = "";
char str[] = "Hello";
s = str;
while(*s)
printf("%c", *s++);

return 0;
}

[A]. Error
[B]. H
[C]. Hello
[D]. Hel

Answer: Option C

Explanation:

Step 1: const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string.

Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".

Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello".

Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s.

Hence the output of the program is "Hello".

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Computer Science & Engineering :: C Programming Questions and Answers : More Exercises

  • Declarations and Initializations
  • Expressions
  • Functions
  • Pointers
  • Strings
  • Input / Output
  • Bitwise Operators
  • Const
  • Variable Number of Arguments
  • Library Functions
  • Control Instructions
  • Floating Point Issues
  • C Preprocessor
  • Arrays
  • Structures, Unions, Enums
  • Command Line Arguments
  • Typedef
  • Memory Allocation
  • Complicated Declarations

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