#include<stdio.h> int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; printf("%u, %u\n", a+1, &a+1); return 0; }
Step 2: printf("%u, %u\n", a+1, &a+1);
The base address(also the address of the first element) of array is 65472.
For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480
Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes".
Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496
Hence the output of the program is 65480, 65496
#include<stdio.h> #include<stdlib.h> union employee { char name[15]; int age; float salary; }; const union employee e1; int main() { strcpy(e1.name, "K"); printf("%s %d %f", e1.name, e1.age, e1.salary); return 0; }
The output will be (in 16-bit platform DOS):
K 75 0.000000#include<stdio.h> int main() { struct bits { int i:40; }bit; printf("%d\n", sizeof(bit)); return 0; }
#include<stdio.h> #include<stdarg.h> int main() { void display(int num, ...); display(4, 12.5, 13.5, 14.5, 44.3); return 0; } void display(int num, ...) { float c; int j; va_list ptr; va_start(ptr, num); for(j=1; j<=num; j++) { c = va_arg(ptr, float); printf("%f", c); } }
/* sample.c */ #include<stdio.h> int main(int argc, char *argv[]) { int j; j = argv[1] + argv[2] + argv[3]; printf("%d", j); return 0; }
Example: j = atoi(argv[1]) + atoi(argv[2]) + atoi(argv[3]);
#include<stdio.h> int main() { float a=3.14; char *j; j = (char*)&a; printf("%d\n", *j); return 0; }
#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 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 x = 10, y = 20; if(!(!x) && x) printf("x = %d\n", x); else printf("y = %d\n", y); return 0; }
Step 1: if(!(!x) && x)
Step 2: if(!(!10) && 10)
Step 3: if(!(0) && 10)
Step 3: if(1 && 10)
Step 4: if(TRUE) here the if condition is satisfied. Hence it prints x = 10.
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