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In C how to print a semicolon without using a semicolon anywhere in the code

print a semicolon without using a semicolon anywhere in the code.

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ha ha tricky question…!!!

#include
int main(void)
{
if (printf("%c
", 59)) //prints the character with ascii value 59, i.e., semicolon
{}

}

Output:

;

Generally when use printf("") statement we have to use semicolon at the end.

If we want to print a semicolon, we use the statement: printf(";");

In above statement, we are using two semicolons. The task of printing a semicolon without using semicolon anywhere in the code can be accomplished by using the ascii value of ’ ; ’ which is equal to 59.

Program: Program to print a semicolon without using semicolon in the code.

#include

int main(void) {

//prints the character with ascii value 59, i.e., semicolon

if (printf("%c
", 59)) {

//prints semicolon

}

return 0;

}

Output:

;

Explanation:

If statement checks whether return value of printf function is greater than zero or not. The return

value of function call printf("%c",59) is 1. As printf returns the length of the string printed. printf("%c",59) prints ascii value that corresponds to 59, that is semicolon(;).