Statements that contain runtime type conversions suffer a little performance penalty each time the statement is executed. If the statement is located in a portion of the program where there is a lot of activity, the total penalty can be significant.

People have their reasons for writing applications with mixed typing. Often it is a matter of saving memory space, memory bandwidth, or time. In the past, for instance, double-precision calculations took twice as long as their single-precision counterparts, so if some of the calculations could be arranged to take place in single precision, there could be a performance win.1 But any time saved by performing part of the calculations in single precision and part in double precision has to be measured against the additional overhead caused by the runtime type conversions. In the following code, the addition of `A(I)`

to `B(I)`

is *mixed type*:

INTEGER NUMEL, I
PARAMETER (NUMEL = 1000)
REAL*8 A(NUMEL)
REAL*4 B(NUMEL)
DO I=1,NUMEL
A(I) = A(I) + B(I)
ENDDO

In each iteration, `B(I)`

has to be promoted to double precision before the addition can occur. You don’t see the promotion in the source code, but it’s there, and it takes time.

C programmers beware: in Kernighan and Ritchie (K&R) C, all floating-point calculations in C programs take place in double precision — even if all the variables involved are declared as *float*. It is possible for you to write a whole K+R application in one precision, yet suffer the penalty of many type conversions.

Another data type–related mistake is to use character operations in IF tests. On many systems, character operations have poorer performance than integer operations since they may be done via procedure calls. Also, the optimizers may not look at code using character variables as a good candidate for optimization. For example, the following code:

DO I=1,10000
IF ( CHVAR(I) .EQ. ’Y’ ) THEN
A(I) = A(I) + B(I)*C
ENDIF
ENDDO

might be better written using an integer variable to indicate whether or not a computation should be performed:

DO I=1,10000
IF ( IFLAG(I) .EQ. 1 ) THEN
A(I) = A(I) + B(I)*C
ENDIF
ENDDO

Another way to write the code, assuming the IFLAG variable was 0 or 1, would be as follows:

DO I=1,10000
A(I) = A(I) + B(I)*C*IFLAG(I)
ENDDO

The last approach might actually perform slower on some computer systems than the approach using the IF and the integer variable.

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