A Newbie Question

[crackedh]

Hi,
I'm a newbie to C, and to programming in general.

My question is:
-----------------------------------------------------------------------------------------------------------------------------------------------
#include <stdio.h>
main()
{
int fahr;
for (fahr = 300; fahr >=0; fahr = fahr - 20)
{
printf("%3.0d %3.2f\n", fahr, (5.0/9.0) * (fahr -
32));
}
}
------------------------------------------------------------------------------------------------------------------------------------------------
#include <stdio.h>
main()
{
int fahr;
for (fahr = 300; fahr >=0; fahr = fahr - 20)
{
printf("%3.0f %3.2f\n", fahr, (5.0/9.0) * (fahr -
32));
}
}

 

[Richard Heathfield]

crackedh said:

Hi,
I'm a newbie to C, and to programming in general.

My question is:
-----------------------------------------------------------------------------------------------------------------------------------------------
#include <stdio.h>
main()
{
int fahr;
for (fahr = 300; fahr >=0; fahr = fahr - 20)
{
printf("%3.0d %3.2f\n", fahr, (5.0/9.0) * (fahr -
32));
}
}
------------------------------------------------------------------------------------------------------------------------------------------------
#include <stdio.h>
main()
{
int fahr;
for (fahr = 300; fahr >=0; fahr = fahr - 20)
{
printf("%3.0f %3.2f\n", fahr, (5.0/9.0) * (fahr -
32));
}
}
--------------------------------------------------------------------------------------------------------------------------------------------------

Why is it that both these segments of code give very diffrent outputs.

The first program looks correct to me (although it is poor style not to
return a value from main - if in doubt, return 0).

The second program asks printf to display a double precision floating-point
number, but provides fahr (an int, not a double) as the value to display.
This breaks the rules of C, and any output you get is therefore arbitrary
- indeed, the behaviour of the whole program becomes undefined.

To display an int using printf, match it with %d. To display a double,
match it with %f.

The second format specifier in that line, %3.2f, is fine, because it's
matched by a double. What's happening here is that fahr is an int and 32
is an int, so fahr - 32 is an int, but then you multiply by a double (that
is, 5.0/9.0), so the (fahr - 32) value is promoted to double for the
purposes of the calculation, and the expression as a whole yields a
double.

 

[Joachim Schmitz]

Hi,
I'm a newbie to C, and to programming in general.

My question is:

int main(void)

{
int fahr;
for (fahr = 300; fahr >=0; fahr = fahr - 20)
{
printf("%3.0d %3.2f\n", fahr, (5.0/9.0) * (fahr -
32))
} return 0
}

int main(void)

{
int fahr;
for (fahr = 300; fahr >=0; fahr = fahr - 20)
{
printf("%3.0f %3.2f\n", fahr, (5.0/9.0) * (fahr -
32));

fahr is int, %f expects double.

}

return 0

}

Because you tell printf to expect a double (%f) but give an int.

Bye, Jojo

 

[crackedh]

Thanks for the exdplanation.

 

[crackedh]

Thanks

 

[Mark Bluemel]

#include <stdio.h>
main()
{
int fahr;
for (fahr = 300; fahr >=0; fahr = fahr - 20)
{
printf("%3.0f %3.2f\n", fahr, (5.0/9.0) * (fahr - 32));

You lied to printf - the "%3.0f" indicated that you would pass it a
double value, and "fahr" is an int.

 

[Martin Ambuhl]

crackedh wrote:
[...]

int fahr; [...]
printf("%3.0f %3.2f\n", fahr, (5.0/9.0) * (fahr -
32));

fahr is an int. The specifier "%3.0f" tells printf that the
corresponding argument (fahr) is a double (or a float whose value is
passed as a double). You lied to printf. Why should you expect the
output to make sense?

 

[CBFalconer]

Thanks for the exdplanation.

Don't answer anything without an explanatory quote. See below.

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