what is the output of the below program & why ?
#include<stdio.h>
void main()
{
int a=10,b=20,c=30;
printf("%d",scanf("%d%d%d",&a,&b,&c));
}
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Answer / gaurav sharma
because printf returns the number of printed arguments .here
3 arguments taken ,the concept is ( scanf also return the
number of arguments so 3 returned to printf .m gaurav
sharma ,,,,,works in aricent.
| Is This Answer Correct ? | 8 Yes | 0 No |
Answer / vibekananda dash
dear selvam,
its showing o/p,
its ask for entry and it accept ur entry,
o/p is -
i entered
1
press enter 2
press enter 78
press enter 3.
my question is why its showing 3.
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / kalyani
The answer is 3 because scanf returns no of arguments
| Is This Answer Correct ? | 5 Yes | 1 No |
for Gcc compiler the code is producing error infact DevC++ just crashed . But using ANSI turbo C++ Ans is 3 ,
As the number of parameters in the printf() function is 3
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / selvam
this program contain no error and it will not give any output...
| Is This Answer Correct ? | 4 Yes | 5 No |
main( ) { char *q; int j; for (j=0; j<3; j++) scanf(“%s” ,(q+j)); for (j=0; j<3; j++) printf(“%c” ,*(q+j)); for (j=0; j<3; j++) printf(“%s” ,(q+j)); }
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1 o 1 1 0 1 0 1 0 1 1 0 1 0 1 how to design this function format in c-language ?
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main() { char *p; p="%d\n"; p++; p++; printf(p-2,300); }
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3 Answers Disney, Google, ZS Associates,
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print a semicolon using Cprogram without using a semicolon any where in the C code in ur program!!
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C Code 
