#include<stdio.h> int main() { char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"}; printf("%d, %d", sizeof(str), strlen(str[0])); return 0; }
Step 2: printf("%d, %d", sizeof(str), strlen(str[0]));
sizeof(str) denotes 6 * 4 bytes = 24 bytes. Hence it prints '24'
strlen(str[0])); becomes strlen(Frogs)). Hence it prints '5';
Hence the output of the program is 24, 5
Hint: If you run the above code in 16 bit platform (Turbo C under DOS) the output will be 12, 5. Because the pointer occupies only 2 bytes. If you run the above code in Linux (32 bit platform), the output will be 24, 5 (because the size of pointer is 4 bytes).
#include<stdio.h> #include<stdlib.h> int main() { union test { int i; float f; char c; }; union test *t; t = (union test *)malloc(sizeof(union test)); t->f = 10.10f; printf("%f", t->f); return 0; }
#include<stdio.h> int main() { int x=4, y, z; y = --x; z = x--; printf("%d, %d, %d\n", x, y, z); return 0; }
Example:
#include <stdio.h>
float sub(float, float); /* Function prototype */
int main()
{
float a = 4.5, b = 3.2, c;
c = sub(a, b);
printf("c = %f\n", c);
return 0;
}
float sub(float a, float b)
{
return (a - b);
}
Output:
c = 1.300000
#include<stdio.h> int main() { int a=100, b=200, c; c = (a == 100 || b > 200); printf("c=%d\n", c); return 0; }
#include<stdio.h> int main() { int x=55; printf("%d, %d, %d\n", x<=55, x=40, x>=10); return 0; }
#include<stdio.h> int main() { void fun(char*); char a[100]; a[0] = 'A'; a[1] = 'B'; a[2] = 'C'; a[3] = 'D'; fun(&a[0]); return 0; } void fun(char *a) { a++; printf("%c", *a); a++; printf("%c", *a); }
#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); }
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