#include<stdio.h> void fun(int); typedef int (*pf) (int, int); int proc(pf, int, int); int main() { int a=3; fun(a); return 0; } void fun(int n) { if(n > 0) { fun(--n); printf("%d,", n); fun(--n); } }
#include<stdio.h> int fun(int); int main() { float k=3; fun(k=fun(fun(k))); printf("%f\n", k); return 0; } int fun(int i) { i++; return i; }
#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 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() { char j=1; while(j < 5) { printf("%d, ", j); j = j+1; } printf("\n"); return 0; }
#include<stdio.h> int main() { int P = 10; switch(P) { case 10: printf("Case 1"); case 20: printf("Case 2"); break; case P: printf("Case 2"); break; } return 0; }
The case statements will accept only constant expression.
#include<stdio.h> #include<math.h> int main() { printf("%f\n", sqrt(36.0)); return 0; }
Declaration Syntax: double sqrt(double x) calculates and return the positive square root of the given number.
#include<stdio.h> int main() { int i = 1; switch(i) { case 1: printf("Case1"); break; case 1*2+4: printf("Case2"); break; } return 0; }
It prints "Case1"
#include<stdio.h> int main() { int i = 10, j = 15; if(i % 2 = j % 3) printf("CuriousTab\n"); return 0; }
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