What does “real*8” mean

bitfortranfortran90precision

The manual of a program written in Fortran 90 says, "All real variables and parameters are specified in 64-bit precision (i.e. real*8)."

According to Wikipedia, single precision corresponds to 32-bit precision, whereas double precision corresponds to 64-bit precision, so apparently the program uses double precision.

But what does real*8 mean?

I thought that the 8 meant that 8 digits follow the decimal point. However, Wikipedia seems to say that single precision typically provides 6-9 digits whereas double precision typically provides 15-17 digits. Does this mean that the statement "64-bit precision" is inconsistent with real*8?

Best Answer

The 8 refers to the number of bytes that the data type uses.

So a 32-bit integer is integer*4 along the same lines.

A quick search found this guide to Fortran data types, which includes:

The "real*4" statement specifies the variable names to be single precision 4-byte real numbers which has 7 digits of accuracy and a magnitude range of 10 from -38 to +38. The "real" statement is the same as "real*4" statement in nearly all 32-bit computers.

and

The "real*8" statement specifies the variable names to be double precision 8-byte real numbers which has 15 digits of accuracy and a magnitude range of 10 from -308 to +308. The "double precision" statement is the same as "real*8" statement in nearly all 32-bit computers.

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