Two decimal places using printf( )

Question

I'm trying to write a number to two decimal places using printf() as follows:

#include <cstdio>
int main()
{
  printf("When this number: %d is assigned to 2 dp, it will be: 2%f ", 94.9456, 94.9456);
  return 0;
}

When I run the program, I get the following output:

# ./printf
When this number: -1243822529 is assigned to 2 db, it will be: 2-0.000000

Why is that?

Thanks.

Answer 1

What you want is %.2f , not 2%f .

Also, you might want to replace your %d with a %f ;)

#include <cstdio>
int main()
{
printf("When this number: %f is assigned to 2 dp, it will be: %.2f ", 94.9456, 94.9456);
return 0;
}

This will output:

When this number: 94.945600 is assigned to 2 dp, it will be: 94.95

See here for a full description of the printf formatting options: printf

Answer 2

Use: "%.2f" or variations on that.

See the POSIX spec for an authoritative specification of the printf() format strings. Note that it separates POSIX extras from the core C99 specification. There are some C++ sites which show up in a Google search, but some at least have a dubious reputation, judging from comments seen elsewhere on SO.

Since you're coding in C++, you should probably be avoiding printf() and its relatives.

Answer 3

For %d part refer to this How does this program work? and for decimal places use %.2f

Answer 4

Try using a format like %d.%02d

int iAmount = 10050;
printf("The number with fake decimal point is %d.%02d", iAmount/100, iAmount%100);

Another approach is to type cast it to double before printing it using %f like this:

printf("The number with fake decimal point is %0.2f", (double)(iAmount)/100);

My 2 cents :)