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Categories  >>  Code Snippets  >>  Programming Code  >>  C Code
 
 


 

 
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Question
how to return a multiple value from a function?
 Question Submitted By :: Guest
I also faced this Question!!     Rank Answer Posted By  
 
  Re: how to return a multiple value from a function?
Answer
# 1
You will have to use a structure or an array and then return
a pointer to the structure/array.
 
Is This Answer Correct ?    1 Yes 1 No
Krishna
 
  Re: how to return a multiple value from a function?
Answer
# 2
Use an array
int [] funct1();
 
Is This Answer Correct ?    2 Yes 0 No
Tanish
 
 
 

 
 
 
Other C Code Interview Questions
 
  Question Asked @ Answers
 
main() { int a[10]; printf("%d",*a+1-*a+3); }  1
main() { int i; clrscr(); printf("%d", &i)+1; scanf("%d", i)-1; } a. Runtime error. b. Runtime error. Access violation. c. Compile error. Illegal syntax d. None of the above HCL1
union u { struct st { int i : 4; int j : 4; int k : 4; int l; }st; int i; }u; main() { u.i = 100; printf("%d, %d, %d",u.i, u.st.i, u.st.l); } a. 4, 4, 0 b. 0, 0, 0 c. 100, 4, 0 d. 40, 4, 0 HCL1
Which version do you prefer of the following two, 1) printf(“%s”,str); // or the more curt one 2) printf(str);  1
main() { int i=10; void pascal f(int,int,int); f(i++,i++,i++); printf(" %d",i); } void pascal f(integer :i,integer:j,integer :k) { write(i,j,k); }  1
typedef struct error{int warning, error, exception;}error; main() { error g1; g1.error =1; printf("%d",g1.error); }  1
main() { int i = 3; for (;i++=0;) printf(“%d”,i); }  1
const int perplexed = 2; #define perplexed 3 main() { #ifdef perplexed #undef perplexed #define perplexed 4 #endif printf("%d",perplexed); } a. 0 b. 2 c. 4 d. none of the above HCL1
main() { char *p="hai friends",*p1; p1=p; while(*p!='\0') ++*p++; printf("%s %s",p,p1); }  1
main() { float me = 1.1; double you = 1.1; if(me==you) printf("I love U"); else printf("I hate U"); }  1
main() { void swap(); int x=10,y=8; swap(&x,&y); printf("x=%d y=%d",x,y); } void swap(int *a, int *b) { *a ^= *b, *b ^= *a, *a ^= *b; }  1
What are the following notations of defining functions known as? i. int abc(int a,float b) { /* some code */ } ii. int abc(a,b) int a; float b; { /* some code*/ }  1
void main() { static int i; while(i<=10) (i>2)?i++:i--; printf(“%d”, i); }  1
main() { int c[ ]={2.8,3.4,4,6.7,5}; int j,*p=c,*q=c; for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){ printf(" %d ",*p); ++p; } }  1
main() { char *p; int *q; long *r; p=q=r=0; p++; q++; r++; printf("%p...%p...%p",p,q,r); }  1
main() { extern int i; i=20; printf("%d",sizeof(i)); }  1
main() { printf("%d, %d", sizeof('c'), sizeof(100)); } a. 2, 2 b. 2, 100 c. 4, 100 d. 4, 4 HCL3
struct Foo { char *pName; char *pAddress; }; main() { struct Foo *obj = malloc(sizeof(struct Foo)); clrscr(); obj->pName = malloc(100); obj->pAddress = malloc(100); strcpy(obj->pName,"Your Name"); strcpy(obj->pAddress, "Your Address"); free(obj); printf("%s", obj->pName); printf("%s", obj->pAddress); } a. Your Name, Your Address b. Your Address, Your Address c. Your Name Your Name d. None of the above HCL1
Is there any difference between the two declarations, 1. int foo(int *arr[]) and 2. int foo(int *arr[2])  1
main(){ char a[100]; a[0]='a';a[1]]='b';a[2]='c';a[4]='d'; abc(a); } abc(char a[]){ a++; printf("%c",*a); a++; printf("%c",*a); }  1
 
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