if x=10 y=10 z=5 how is x

问题

I have this question & in the answer it says that due to left to right associativity the result would be 1 that is true for this statement. This is the code.

#include<stdio.h>

int main ()
{
int i=0,x=10,y=10,z=5;
i=x<y<z;
printf("\n\n%d",i);

return 0;
}

But x is greater than z here so how is this happening ?

回答1:

The expression x

(x < y) < z

so it becomes

(10 < 10) < 5

which further is evaluated into

0 < 5

which is true.

I think you wanted something like this:

x < y && y < z

回答2:

Because of operator precedence and associativity

i = x < y < z;

is parsed as:

i = ((x < y) < z);

After substituting the variable values, this becomes:

i = ((10 < 10) < 5);

Since 10 < 10 is false, this becomes:

i = (0 < 5);

Since 0 < 5 is true, that becomes:

i = 1;

回答3:

x<y<z is not a single valid expression. Instead it evaluates x<y first (operator precedence is done left to right here) as true/false (false in this case as they're equal), converts it to an int value of 0, and then compares this value with z.

Use (x < y && y < z) instead.

回答4:

It first evaluates x < y which is false (0), then 0 < z which is true (1).

回答5:

WHat C compiler does is, in x<y<z; starts from left, so as x is not less than y therefore it replaces that expression with '0' so it becomes 0<z and as that is true. it set the variable to 1.

来源：https://stackoverflow.com/questions/21202683/if-x-10-y-10-z-5-how-is-xyz-true

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