FIND THE OUTPUT IF THE INPUT IS 5 5.75 void main() { int i=1; float f=2

FIND THE OUTPUT IF THE INPUT IS 5 5.75
void main()
{
int i=1;
float f=2.25;
scanf("%d%f",&i,&f);
printf("%d %f",,i,f);
}
ANSWER IS 5 AND 2.25 WHY?

Question Posted / biren

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FIND THE OUTPUT IF THE INPUT IS 5 5.75 void main() { int i=1; float f=2.25; scanf("..

Answer / biren

5 2.25

Is This Answer Correct ?

6 Yes

4 No

FIND THE OUTPUT IF THE INPUT IS 5 5.75 void main() { int i=1; float f=2.25; scanf("..

Answer / hari.11

hey,
if we give input as 5 5.75
output would be: 5 5.750000

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Is This Answer Correct ?

1 Yes

0 No

FIND THE OUTPUT IF THE INPUT IS 5 5.75 void main() { int i=1; float f=2.25; scanf("..

Answer / suman halder

There is a tiny mistake in the question..It should be..

FIND THE OUTPUT IF THE INPUT IS 5 5.75

void main()
{
int i=1;
float f=2.25;
scanf("%d %d",&i,&f);
printf("%d %f",,i,f);
}
ANSWER IS 5 AND 2.25 WHY?

The answer of this question actually lies within the question itself..
if we look at the scanf statement ,then the actual reason of the output will be revealed..
In scanf,we are trying to get the input into a floating point number through %d format specifier which is the format specifier for integers..here,lies the anomaly..
nothing actually get stored within f through scanf ,so the previous value of f is retained..and that gets printed..
thats it..

*** if we try the reverse of it.i.e using %f,we wanna get the value into an integer..the compiler may report some error like "floating point format not linked"...we can abstain the error by intoducing a linkfloat function but still there will be some anomaly in the value that get stored in the variable ..so,to get the expected output we need to specify the correct format specifier....dnt trust printf or scanf to perform the proper typecasting for u..
do it explicitly..

linkfloat()
{
float a,*b;
b=&a;
a=*b;
}

Is This Answer Correct ?

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FIND THE OUTPUT IF THE INPUT IS 5 5.75 void main() { int i=1; float f=2.25; scanf("..

Answer / reshma pawar

there is error in
printf("%d %f",,i,f);
because 2 (,) operators are used
an if operator is removed then output will be
5 5.750000

Is This Answer Correct ?

0 Yes

0 No

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