Answer / susie

Answer :

1

Explanation:

before entering into the for loop the checking condition is
"evaluated". Here it evaluates to 0 (false) and comes out of
the loop, and i is incremented (note the semicolon after the
for loop).

main() { extern int i; i=20; printf("%d",sizeof(i)); }

2 Answers  

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main() { int x=5; clrscr(); for(;x==0;x--) { printf("x=%d\n”", x--); } } a. 4, 3, 2, 1, 0 b. 1, 2, 3, 4, 5 c. 0, 1, 2, 3, 4 d. none of the above

3 Answers   HCL,

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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”); }

1 Answers  

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main(int argc, char **argv) { printf("enter the character"); getchar(); sum(argv[1],argv[2]); } sum(num1,num2) int num1,num2; { return num1+num2; }

1 Answers  

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

2 Answers  

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func(a,b) int a,b; { return( a= (a==b) ); } main() { int process(),func(); printf("The value of process is %d !\n ",process(func,3,6)); } process(pf,val1,val2) int (*pf) (); int val1,val2; { return((*pf) (val1,val2)); }

1 Answers   Satyam,

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main() { int i, j, *p; i = 25; j = 100; p = &i; // Address of i is assigned to pointer p printf("%f", i/(*p) ); // i is divided by pointer p } a. Runtime error. b. 1.00000 c. Compile error d. 0.00000

3 Answers   HCL,

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x=2 y=3 z=2 x++ + y++; printf("%d%d" x,y);

2 Answers  

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How will you print % character? a. printf(“\%”) b. printf(“\\%”) c. printf(“%%”) d. printf(“\%%”)

4 Answers   HCL,

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programming in c lanugaue programm will errror error with two header file one as stdio.h and other one is conio.h

1 Answers  

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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

1 Answers   emc2, HCL,

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main() { int i = 100; clrscr(); printf("%d", sizeof(sizeof(i))); } a. 2 b. 100 c. 4 d. none of the above

5 Answers   HCL,

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