Looking at the code:
int i = 5;
if (i = 0)
{
printf ("Got here\n");
}
What does the C standard have to say about what will get printed? Or in more general terms does the assignment happen first or the comparison?
Looking at the code:
int i = 5;
if (i = 0)
{
printf ("Got here\n");
}
What does the C standard have to say about what will get printed? Or in more general terms does the assignment happen first or the comparison?
The assignment happens, which returns a 0, which is false.
Assignment first, as it is part of the evaluation. The expression of the assignment returns the assigned value, so the expression evaluates as false.
The expression of the if clause is evaluated first, the result of which is 0.
This program will never print "Got here\n".
i=0
evaluates to 0 thus the output will not happen.
The prior assignment (the first line of source code) is irrelevant to the outcome.
The statement i = 0
will be evaluated and return 0
, so the statement will not be printed.
As others already said, the assignment returns the value of that is assigned and so never prints the statement. If you wanted the statement to be printed, you'd have to use if (i = -1)
.
Nothing will print. The 0 get assigned to i and then that value is tested for the condition.
When the assignment happens is irrelevant. What's relevant is the value of i=0
as an expression, and it's defined to have the value 0.