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what is op?
for(c=0;c=1000;c++)
printf("%c",c);
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Answer / vikas chauhan
It will print funny caharcter ( ascii equivalent of 1000
trucated char) infinitely
| Is This Answer Correct ? | 14 Yes | 3 No |
Answer / karthik
i executed this program it went on displaying funny
characters indefinite loop
| Is This Answer Correct ? | 8 Yes | 1 No |
Answer / abdur rab
It goes into an infinite loop.
The compiler will check only the syntax, it is upto the
programmer to implement correct logic.
other than initialization, c is always 1000 and there is no
condition to be break the loop
| Is This Answer Correct ? | 5 Yes | 0 No |
Answer / sriharsha
It print's only what's the ASCII value of 1000 for infinite loop
| Is This Answer Correct ? | 4 Yes | 2 No |
Answer / dally
it will goto infinite loop because each time i=0 and i=100
so it will goto infinite loop.
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / sreekanth
Warnings:
Test expression for for is assignment expression: c = 1000
The condition test is an assignment expression. Probably,
you mean to use == instead of =. If an assignment is
intended, add an extra parentheses nesting(e.g., if ((a =
b)) ...) to suppress this message. (Use -predassign to
inhibit warning)
Test expression for for not boolean, type int: c = 1000.
Test expression type is not boolean or int. (Use
-predboolint to inhibit warning)
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / m.kiran kumar
warning occurs as possibly incorrect statement.
and the statement falls into infinite loop printing the
ascii character equivalent to 0
| Is This Answer Correct ? | 2 Yes | 2 No |
What is the output of the following progarm? #include<stdio.h> main( ) { int x,y=10; x=4; y=fact(x); printf(“%d\n”,y); } unsigned int fact(int x) { return(x*fact(x-1)); } A. 24 B. 10 C. 4 D. none
main() { int i; printf("%d",i^i); }
write the output of following code .. main() { static int a[]={10,20,30,40,50}; int *ptr=a; static int arr[2][2]={1,2,3,4}; char str[]="ABCD * 4#"; char *s=str+2; int i,j; for(i=0;i<5,i++) printf("%d",*ptr++); for(i=0;i<2;i++) for(j=0;j<2;j++) printf("%d\n",*(*(n+i)+j)); printf("%c\n%c\n%c\n",*(str+2),*s++,*--s); }
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