assigning a large float number in E notation

edM · 05-09-2005, 11:17 AM

hi,

how do you assign a large flatoing point number in C, is the syntax below correct?

float number1 = 2.302E-12; //delcare & assign
.
.
.
.
printf("%f", number1);

when i went to print it out all i got was '0.000000'

thanks

enemorales · 05-09-2005, 12:03 PM

Code:

 float number1 = 2.302E-12

Well, this is actually a very SMALL number: about 0.000000000002302, to be more precise. This is because you put a MINUS (E-12)

. Did you mean 2.302E12? (Which i think it is also 2.302E+12)

Hko · 05-09-2005, 12:03 PM

Quote:

Originally posted by edM
float number1 = 2.302E-12; //delcare & assign
[...]
printf("%f", number1);

when i went to print it out all i got was '0.000000'

printf()'s defaults to 6 digits after the decimal point for "%f " conversion. And 2.302E-12 is smaller than 0.000000, so it's rounded to zero.

Either printf if in E-notatation:

Code:

printf("%E\n", number1);

Or specify how much digits after the decimal point should be output (precision), e.g. to get 15 digits:

Code:

printf("%.15f\n", number1);

edM · 05-09-2005, 12:51 PM

enemorales yes i meant large as in small.

thanks both you, very helpful!