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