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关闭调试,NDEBUG

[英]Turning off debugging, NDEBUG

C++ Primer says that C ++ Primer说

The behavior of assert depends on the status of a preprocessor variable named NDEBUG. assert的行为取决于名为NDEBUG的预处理程序变量的状态。 We can "turn off" debugging by providing a #define to define NDEBUG 我们可以通过提供#define定义NDEBUG来“关闭”调试

It is my expectation that when define is provided, asserts won't be executed. 我期望当提供define时,断言将不会被执行。

#define NDEBUG TRUE

int main (int argc, char const *argv[])
{
    assert(argc==0);  // checked

    return 0;
}

Why, in this example, is assert statement checked, when NDEBUG is defined? 在此示例中,为什么在定义NDEBUG时检查assert语句? (Correct me if i am wrong, but it does not matter to what it is defined, right?) (如果我错了,请更正我,但是定义的内容无关紧要,对吧?)

When executed from command line, using the -DNDEBUG flag, all works as expected ( assert is not executed) 使用-DNDEBUG标志从命令行执行时,所有操作-DNDEBUG预期进行(未执行assert

NDEBUG only affects assert if you define it before including <cassert> (or <assert.h> ; note that you can include these headers multiple times changing the behaviour of assert depending on NDEBUG ). NDEBUG仅在包含<cassert> (或<assert.h>之前定义assert时会影响assert 。请注意,可以多次包含这些标头,从而根据NDEBUG改变assert的行为。

You don't need to define it to any specific value, or any value at all: 您无需将其定义为任何特定值,或者根本不需要将其定义为:

// this is OK
#define NDEBUG

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