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I came across this program via a quora answer
#include<stdio.h>
int main() {
printf("%d\n", ( { int n; scanf("%d", &n); n*n; } ));
return 0;
}
I was wondering how does this work and if this conforms the standard?
256
scanf("%d",&n),n. The value of the condition is the value of the second operand of the comma operator that is of the variable n . If n is not equal to 0 then the loop is executed. You can imagine it the following way start: scanf("%d",&n); if ( n != 0 ) { //...
Description. The scanf() function reads data from the standard input stream stdin into the locations given by each entry in argument-list. Each argument must be a pointer to a variable with a type that corresponds to a type specifier in format-string.
In C programming, scanf() is one of the commonly used function to take input from the user. The scanf() function reads formatted input from the standard input such as keyboards.
scanf requires the addressOf operator (&) because it takes a pointer as an argument. Therefore in order to pass in a variable to be set to a passed in value you have to make a pointer out of the variable so that it can be changed.
This code is using a "GNU C" feature called statement-expressions, whereby a parentheses-enclosed compound statement can be used as an expression, whose type and value match the result of the last statement in the compound statement. This is not syntactically valid C, but a GCC feature (also adopted by some other compilers) that was added presumably because it was deemed important for writing macros which do not evaluate their arguments more than once.
You should be aware of what it is and what it does in case you encounter it in code you have to read, but I would avoid using it yourself. It's confusing, unnecessary, and non-standard. The same thing can almost always be achieved portably with static inline functions.
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