目前 C++的装运时间为malloc(>>,与http://msdn.microsoft.com/en-us/library/8bcxafdh%28v=vs.80%29.aspx” rel=“nofollow”>_declspec ( limit )<<><>>>>>>>>>。
MSDN说,这种校正向汇编者说,a pointer用malloc(<>>/code>返回,不能由任何其他点名加以补充。 Okay, 2个后来的电话:malloc()的确是返回的单独点。 但是,如果我呼吁,情况会发生。
void* memory1 = malloc( 10 );
free( memory1 );
void* memory2 = malloc( 10 );
//here memory1 may be equal to memory2
在这种情况下,两点人可以指出同一地点。 这一点如何与相挂钩,无法由任何其他点名<>/m>内含<代码>-declspec(限制?
The standard has this to say about "object lifetime" (§3.8 in N3290):
The lifetime of an object of type T ends when:
— if T is a class type with a non-trivial destructor (12.4), the destructor call starts, or
— the storage which the object occupies is reused or released.
在您通过<条码>免费后,该物体已亡故。 它已经不复存在。 忽略该点人是不明确的行为。
因为一旦你释放(主题1),通过记忆1点人获取任何东西都是不明确的行为(新陈词,等等),因此,汇编者可以优化假设,在小鼠打电话后,其他任何点人不会把记忆2混为一谈。
因此,假定汇编者自己没有内部资料,说明小型(小型)的属性,即它只处理它与其他任何职能一样,则不能假定被送回的点子不是由任何其他点人所决定的。 declspec(restrict) (或等同,attribute__(malloc)>, inlberg) 告知汇编者,该点人不是由任何其他点人挑选的,因此无法以其他方式加以优化。
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