#include<stdio.h> int addmult(int ii, int jj) { int kk, ll; kk = ii + jj; ll = ii * jj; return (kk, ll); } int main() { int i=3, j=4, k, l; k = addmult(i, j); l = addmult(i, j); printf("%d %d\n", k, l); return 0; }
#include<stdio.h> #define MESS junk int main() { printf("MESS\n"); return 0; }
#include<stdio.h> int main() { static char *s[] = {"black", "white", "pink", "violet"}; char **ptr[] = {s+3, s+2, s+1, s}, ***p; p = ptr; ++p; printf("%s", **p+1); return 0; }
#include<stdio.h> int main() { int arr[3] = {2, 3, 4}; char *p; p = arr; p = (char*)((int*)(p)); printf("%d, ", *p); p = (int*)(p+1); printf("%d", *p); return 0; }
1 : | typedef long a; extern int a c; |
2 : | typedef long a; extern a int c; |
3 : | typedef long a; extern a c; |
typedef long a;
extern a int c; while compiling this statement becomes extern long int c;. This will result in to "Too many types in declaration error".
typedef long a;
extern a c; while compiling this statement becomes extern long c;. This is a valid c declaration statement. It says variable c is long data type and defined in some other file or module.
So, Option C is the correct answer.
char int; here int is a keyword cannot be used a variable name.
int long; here long is a keyword cannot be used a variable name.
float double; here double is a keyword cannot be used a variable name.
So, the answer is int length;(Option A).
#include<stdio.h> int main() { int i=2; int j = i + (1, 2, 3, 4, 5); printf("%d\n", j); return 0; }
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