Questions for: Expressions
#include<stdio.h>
int main()
{
int k, num=30;
k = (num>5 ? (num <=10 ? 100 : 200): 500);
printf("%d\n", num);
return 0;
}
Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to '30'.
Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement.
Step 3: printf("%d\n", num); It prints the value of variable num '30'
Step 3: Hence the output of the program is '30'
#include<stdio.h>
int main()
{
int i=2;
printf("%d, %d\n", ++i, ++i);
return 0;
}
The order of evaluation of arguments passed to a function call is unspecified.
Anyhow, we consider ++i, ++i are Right-to-Left associativity. The output of the program is 4, 3.
In TurboC, the output will be 4, 3.
In GCC, the output will be 4, 4.
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#include<stdio.h>
int main()
{
int x=55;
printf("%d, %d, %d\n", x<=55, x=40, x>=10);
return 0;
}
Step 1: int x=55; here variable x is declared as an integer type and initialized to '55'.
Step 2: printf("%d, %d, %d\n", x<=55, x=40, x>=10);
In printf the execution of expressions is from Right to Left.
here x>=10 returns TRUE hence it prints '1'.
x=40 here x is assigned to 40 Hence it prints '40'.
x<=55 returns TRUE. hence it prints '1'.
Step 3: Hence the output is "1, 40, 1".
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#include<stdio.h>
int main()
{
int a=100, b=200, c;
c = (a == 100 || b > 200);
printf("c=%d\n", c);
return 0;
}
Step 1: int a=100, b=200, c;
Step 2: c = (a == 100 || b > 200);
becomes c = (100 == 100 || 200 > 200);
becomes c = (TRUE || FALSE);
becomes c = (TRUE);(ie. c = 1)
Step 3: printf("c=%d\n", c); It prints the value of variable i=1
Hence the output of the program is '1'(one).
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#include<stdio.h>
int main()
{
int i=3;
i = i++;
printf("%d\n", i);
return 0;
}
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