Short circuit behavior of logical operators
suggest changeShort circuiting is a functionality that skips evaluating parts of a (if/while/…) condition when able. In case of a logical operation on two operands, the first operand is evaluated (to true or false) and if there is a verdict (i.e first operand is false when using &&, first operand is true when using ||) the second operand is not evaluated.
Example:
#include <stdio.h>
int main(void) {
int a = 20;
int b = -5;
/* here 'b == -5' is not evaluated,
since a 'a != 20' is false. */
if (a != 20 && b == -5) {
printf("I won't be printed!\n");
}
return 0;
}
Check it out yourself:
#include <stdio.h>
int print(int i) {
printf("print function %d\n", i);
return i;
}
int main(void) {
int a = 20;
/* here 'print(a)' is not called,
since a 'a != 20' is false. */
if (a != 20 && print(a)) {
printf("I won't be printed!\n");
}
/* here 'print(a)' is called,
since a 'a == 20' is true. */
if (a == 20 && print(a)) {
printf("I will be printed!\n");
}
return 0;
}
Output:
$ ./a.out
print function 20
I will be printed!
Short circuiting is important, when you want to avoid evaluating terms that are (computationally) costly. Moreover, it can heavily affect the flow of your program like in this case: http://stackoverflow.com/questions/26716255/why-does-this-program-print-forked-4-times
Found a mistake? Have a question or improvement idea?
Let me know.
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