format command of C that will not round up the value

i have a question
please refer below..

Example:

Code:

---

#include <stdio.h>
int main() {
    printf("%6.1f", 123456.123456);
    printf("\n%6.2f", 123456.123456);
    printf("\n%6.3f", 123456.123456);
    printf("\n%6.4f", 123456.123456);
}

---

output :
123456.1
123456.12
123456.123
123456.1235

at last output the result was 123456.1235 because it was rounded up.

How to display this value by not rounding up?
like 123456.1234 ?

Printf is supposed to round it. To prevent this, truncate the argument to 4 decimal places.

You could sprintf() into a temp buffer for 5 decimal places and then lop off the last digit.

Or, you could multiply the source number by 10,000, cast it to an int, and then back to a float and divide by 10,000.

I'm sure I already answered this question somewhere else...

--
Mats

Quote:

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Originally Posted by Todd Burch

You could sprintf() into a temp buffer for 5 decimal places and then lop off the last digit.

---

could be better but i have no idea to lop it off.
in VB we can used MID() function
but how about in C ?

Quote:

---

Originally Posted by opaw

i have a question
please refer below..

Example:

Code:

---

#include <stdio.h>
int main() {
    printf("%6.1f", 123456.123456);
    printf("\n%6.2f", 123456.123456);
    printf("\n%6.3f", 123456.123456);
    printf("\n%6.4f", 123456.123456);
}

---

output :
123456.1
123456.12
123456.123
123456.1235

at last output the result was 123456.1235 because it was rounded up.

How to display this value by not rounding up?
like 123456.1234 ?

---

Code:

---

#include "stdafx.h"


#include <stdio.h>
int main()
{
    printf("%6.4f", 123456.123456);
    printf("\n%6.4f", 123456.123456);
    printf("\n%6.4f", 123456.123456);
    printf("\n%6.4f", 123456.123456);

        return 0;
}

---

Quote:

---

Originally Posted by opaw

could be better but i have no idea to lop it off.
in VB we can used MID() function
but how about in C ?

---

Code:

---

char test[] = "123456.123456";

test[11] = '\0';

---

perhaps? :)

Of course that's just an example, strchr() may help you.

or a bit of casting:

Code:

---

#include <stdio.h>

int main(void)
{
    double test = 123456.123456;
    int places = 10 * 10 * 10 * 10,    /* 4 decimal places, 10^4 */
        cast = 0,
        test2 = 0;

    /* truncate decimals */
    cast = (int) test;

    test2 = (test - cast) * places;
   
    printf("&#37;d.%d\n", cast, test2);
    return 0;
}

---

Again, it's a demo don't use it :)

Quote:

---

Originally Posted by opaw

could be better but i have no idea to lop it off.
in VB we can used MID() function
but how about in C ?

---

It is an error in itself to compare VB to C. They are nothing alike and they work nothing alike.
If you facilities closer to VB, then C++ is your territory. Otherwise you will have to stick to non-objected approaches and functions and low-level stuff.

Good luck.

Why is everyone making this sound so difficult?

printf("\n%6.4f", (int)(123456.123456*10000)/10000.0);

Quote:

---

Originally Posted by R.Stiltskin

Why is everyone making this sound so difficult?

printf("\n%6.4f", (int)(123456.123456*10000)/10000.0);

---

There's an echo in here...
There's an echo in here...

Quote:

---

Originally Posted by Todd

Or, you could multiply the source number by 10,000, cast it to an int, and then back to a float and divide by 10,000.

---

Quote:

---

Why is everyone making this sound so difficult?

printf("\n%6.4f", (int)(123456.123456*10000)/10000.0);

---

Have you considered the possibility of integer overflow?

If you print it to 5 places, the round might travel through the whole number. You would have to print the whole number at full precision to a buffer and then truncate it, I think.