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need explanation

void main() { static int i=5; if(--i){ main(); printf("%d ",i); } }

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#define a 10 void foo() { #undef a #define a 50 } int main() { printf("%d..",a); foo(); printf("%d..",a); return 0; } explain the answer

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main( ) { static int a[ ] = {0,1,2,3,4}; int *p[ ] = {a,a+1,a+2,a+3,a+4}; int **ptr = p; ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *++ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); ++*ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); }

2 Answers   Persistent,

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main() { int c = 5; printf("%d", main||c); } a. 1 b. 5 c. 0 d. none of the above

2 Answers   HCL,

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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; }

2 Answers  

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main() { char c=' ',x,convert(z); getc(c); if((c>='a') && (c<='z')) x=convert(c); printf("%c",x); } convert(z) { return z-32; }

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What is the problem with the following code segment? while ((fgets(receiving array,50,file_ptr)) != EOF) ;

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main() { char name[10],s[12]; scanf(" \"%[^\"]\"",s); } How scanf will execute?

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Is there any difference between the two declarations, 1. int foo(int *arr[]) and 2. int foo(int *arr[2])

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main() { if (!(1&&0)) { printf("OK I am done."); } else { printf("OK I am gone."); } } a. OK I am done b. OK I am gone c. compile error d. none of the above

3 Answers   HCL,

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#include<stdio.h> main() { struct xx { int x; struct yy { char s; struct xx *p; }; struct yy *q; }; }

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void main() { if(~0 == (unsigned int)-1) printf(“You can answer this if you know how values are represented in memory”); }

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