#include<stdio.h> int main() { int a = 300, b, c; if(a >= 400) b = 300; c = 200; printf("%d, %d, %d\n", a, b, c); return 0; }
#include<stdio.h> int main() { int a = 500, b = 100, c; if(!a >= 400) b = 300; c = 200; printf("b = %d c = %d\n", b, c); return 0; }
Step 1: if(!a >= 400)
Step 2: if(!500 >= 400)
Step 3: if(0 >= 400)
Step 4: if(FALSE) Hence the if condition is failed.
Step 5: So, variable c is assigned to a value '200'.
Step 6: printf("b = %d c = %d\n", b, c); It prints value of b and c.
Hence the output is "b = 100 c = 200"
#include<stdio.h> int main() { int i=4; switch(i) { default: printf("This is default\n"); case 1: printf("This is case 1\n"); break; case 2: printf("This is case 2\n"); break; case 3: printf("This is case 3\n"); } return 0; }
In default statement there is no break; statement is included. So it prints the case 1 statements. "This is case 1".
Then the break; statement is encountered. Hence the program exits from the switch-case block.
#include<stdio.h> int main() { int x = 3; float y = 3.0; if(x == y) printf("x and y are equal"); else printf("x and y are not equal"); return 0; }
#include<stdio.h> int main() { int a=0, b=1, c=3; *((a)? &b : &a) = a? b : c; printf("%d, %d, %d\n", a, b, c); return 0; }
Step 2: *((a) ? &b : &a) = a ? b : c; The right side of the expression(a?b:c) becomes (0?1:3). Hence it return the value '3'.
The left side of the expression *((a) ? &b : &a) becomes *((0) ? &b : &a). Hence this contains the address of the variable a *(&a).
Step 3: *((a) ? &b : &a) = a ? b : c; Finally this statement becomes *(&a)=3. Hence the variable a has the value '3'.
Step 4: printf("%d, %d, %d\n", a, b, c); It prints "3, 1, 3".
#include<stdio.h> int main() { int k, num = 30; k = (num < 10)? 100 : 200; printf("%d\n", num); return 0; }
#include<stdio.h> int main() { char str[]="C-program"; int a = 5; printf(a >10?"Ps\n":"%s\n", str); return 0; }
if(a > 10)
{
printf("Ps\n");
}
else
{
printf("%s\n", str);
}
Here we are checking a > 10 means 5 > 10. Hence this condition will be failed. So it prints variable str.
Hence the output is "C-program".
#include<stdio.h> int main() { char j=1; while(j < 5) { printf("%d, ", j); j = j+1; } printf("\n"); return 0; }
#include<stdio.h> int main() { int i=3; switch(i) { case 1: printf("Hello\n"); case 2: printf("Hi\n"); case 3: continue; default: printf("Bye\n"); } return 0; }
#include<stdio.h> int main() { int x=1, y=1; for(; y; printf("%d %d\n", x, y)) { y = x++ <= 5; } printf("\n"); return 0; }
#include<stdio.h> int main() { int i=1; for(;;) { printf("%d\n", i++); if(i>10) break; } return 0; }
Hence the output of the program is
1
2
3
4
5
6
7
8
9
10
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.