#include<stdio.h> #define CUBE(x) (x*x*x) int main() { int a, b=3; a = CUBE(b++); printf("%d, %d\n", a, b); return 0; }
Step 1: int a, b=3; The variable a and b are declared as an integer type and varaible b id initialized to 3.
Step 2: a = CUBE(b++); becomes
=> a = b++ * b++ * b++;
=> a = 3 * 3 * 3; Here we are using post-increement operator, so the 3 is not incremented in this statement.
=> a = 27; Here, 27 is store in the variable a. By the way, the value of variable b is incremented by 3. (ie: b=6)
Step 3: printf("%d, %d\n", a, b); It prints the value of variable a and b.
Hence the output of the program is 27, 6.
#include<stdio.h> void fun(void *p); int i; int main() { void *vptr; vptr = &i; fun(vptr); return 0; } void fun(void *p) { int **q; q = (int**)&p; printf("%d\n", **q); }
#include<stdio.h> #include<string.h> int main() { static char s[] = "Hello!"; printf("%d\n", *(s+strlen(s))); return 0; }
#include<stdio.h> int main() { int i=4, j=8; printf("%d, %d, %d\n", i|j&j|i, i|j&&j|i, i^j); return 0; }
#include<stdio.h> int main() { int i; printf("%d\n", scanf("%d", &i)); return 0; }
printf("%d\n", scanf("%d", &i)); The scanf function returns the value 1(one).
Therefore, the output of the program is '1'.
int *ptr[30];
#include<stdio.h> int main() { char huge *near *far *ptr1; char near *far *huge *ptr2; char far *huge *near *ptr3; printf("%d, %d, %d\n", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3)); return 0; }
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