I am trying to have a user input a number, 5 for example, and then prints a number i have with that many decimal points.
example:
user types: 5
prints: 1.10293
(i chose random numbers here, but has 5 decimal points)
I only know how to do this printf("%.5lf", mynumber);
and that would print 5, but always 5.
i want it to print the amount the user puts in.
thank you for any help or advice
Last edited by frezzy; 09-27-2017 at 09:53 PM.
There's sprintf that prints whatever you want into a string. Print the number you want, along with the format of course, into a string and pass it as the format string of printf.
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hello thanks, I am sorry but I am very new at this and do not understand.Originally Posted by GReaper
sprintf - C++ Reference
Use it kind-of like this:
Code:char buffer[10]; sprintf(buffer, "%%.%df", digits); // This produces "%.Nf", where N stands for the value held by 'digits' printf(buffer, mynumber); // This takes the string produced above as a format
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@frezzy
I think you were looking for this:
Use the '*' char to fill in the field width of the decimal portion in the format string, input by the user.Code:#include <stdio.h> #include <stdlib.h> int main(int argc, char *argv[]) { double val = 12.3456789; int places = 0; if(argc != 2) { printf("USAGE: %s [decimal place int]\n", argv[0]); printf("Example: %s 2 (For two decimal places)\n", argv[0]); exit(EXIT_FAILURE); } else { places = atoi(argv[1]); } printf("%d Decimal places: %.*lf\n", places, places, val); return 0; }
Oops, completely forgot about that specifier. Use @rstanley's solution, it's better for what you want to do.
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