How to print a semicolon(;) without using semicolon in C/C++?
Another interesting question is how can a semicolon be printed without using any semicolon in the program. Here are methods to print “;” :
- Using printf / putchar in if statement:
CPP
// CPP program to print // ; without using ; // using if statement #include <stdio.h> int main() { // ASCII value of semicolon = 59 if ( printf ("%c\n", 59)) if ( putchar (59)) { } return 0; } |
Output:
; ;
- We can also print the semicolon using switch/ while / for as illustrated in this article.
- Using macros :
CPP
// CPP program to print // ; without using ; // using macros #include <stdio.h> #define GEEK printf("%c",59) int main() { if (GEEK) { } } |
Output:
;
3. Using std::cout
C++
#include <iostream> // 59 is an ASCII value of the semicolon #define SEMICOLON 59 int main() { if (std::cout << static_cast < char >(SEMICOLON)) { } } |
Output
;
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