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print a semicolon using Cprogram without using a semicolon
any where in the C code in ur program!!
- 35 Answers
- 57419 Views
- Tata Elxsi, TCS, VI eTrans, I also Faced
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Answers were Sorted based on User's Feedback
Answer / arunabha
#include<stdio.h>
#include<conio.h>
void main()
{
if(printf("%c",59)){}
if(getch()){}
}
this is the perfect solution...
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / rahul
void main()
{
if(printf("%c",59))
{}
if(getch())
}
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / poorna
#include<stdio.h>
void main()
{
if(printf("\n welcome \n"))
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / neeraj gupta
#include<stdio.h>
void main()
{
printf("%c",59);
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / abinash
void main()
{
if(printf("semicolon")){}
}
this works,if u include (stdio.h)
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / gaurav sharma
void main()
{
if(printf("%c",59))
{}
if(getch())
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / sam prasanna
we can use the ASCII value of semicolon(;) to print it in if condition.
| Is This Answer Correct ? | 0 Yes | 0 No |
void pascal f(int i,int j,int k) { printf(“%d %d %d”,i, j, k); } void cdecl f(int i,int j,int k) { printf(“%d %d %d”,i, j, k); } main() { int i=10; f(i++,i++,i++); printf(" %d\n",i); i=10; f(i++,i++,i++); printf(" %d",i); }
main() { int i=0; while(+(+i--)!=0) i-=i++; printf("%d",i); }
9 Answers CSC, GoDB Tech, IBM,
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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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C Code 
