what do the compiler options mean? and '?'followed by '=-03'

Question

/redis/deps/hiredis/Makefile

    CC:=$(shell sh -c 'type $(CC) >/dev/null 2>/dev/null && echo $(CC) ||      echo gcc')
    OPTIMIZATION?=-O3
    WARNINGS=-Wall -W -Wstrict-prototypes -Wwrite-strings
    DEBUG?= -g -ggdb

what do the complier options mean? and '?'followed by '=-03'?

score 2 · Answer 1 · answered Jan 20 '16 at 07:55

2

for OPTIMIZATION?=-O3

If the variable OPTIMIZATION has NOT been defined, OPTIMIZATION is valued as -O3;
On the other side, if the variable OPTIMIZATION has been defined, this statement will be skipped, and OPTIMIZATION will keep its previous value

answered Jan 20 '16 at 07:55

kaitian521

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Maybe clarify that `?=` is a conditional assignment operator in GNU Make. – tripleee Jan 20 '16 at 12:28

score 0 · Answer 2 · answered Jul 30 '17 at 12:22

The GNU make manual has this text for the ?= operator

There is another assignment operator for variables, ‘?=’. This is called a conditional variable assignment operator, because it only has an effect if the variable is not yet defined.

So, this statement

OPTIMIZATION?=-O3

says that if OPTIMIZATION has been defined, ignore it and if not, define it and assign a value -03.

The -03 as compiler flags means compile at optimization level 3.

what do the compiler options mean? and '?'followed by '=-03'

2 Answers2