对16位整数进行奇怪的按位运算

Question

i'm looking at ac source file and i found this macro:

#define random ( (float) rand() / (float) ((1 << 31) -1) )

while in standard ANSI C rand() returns an integer in [0,32767], i really appreciate an help to understand what kind of normalization factor is the denominator, because signed integer are 16 bit and the expression does a 31-bit shift.

Thank you very much for your attention Best regards

Answer 1

rand does not return an integer in [0,32767] in "ANSI C". §7.20.2:

The rand function computes a sequence of pseudo-random integers in the range 0 to RAND_MAX .

It seems likely that whoever wrote that macro was working on a platform on which RAND_MAX was 2147483647.

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You also seem to be confused about signed integers. int must be at least 16 bits wide, but it is often wider.

Answer 2

#define random ( (float) rand() / (float) ((1 << 31) -1) )

in a system with 16-bit int , this macro is undefined behavior because of 1 << 31 expression ( 1 is of int type).

(C99, 6.5.7p3) "If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand , the behavior is undefined."