[英]iterate char** why does this work?
I picked up a this piece of code I copy past to my program. 我拿起这段代码,将其复制到程序中。 This seems to be a new way to me to iterate through char**: 这似乎是我遍历char **的一种新方法:
char** vArray; // The array containing values
// Go throught properties
if(szKey == "KeyMgmt")
{
vArray = (char**)g_value_get_boxed((GValue*)value);
for( ; vArray && *vArray ; vArray++) // Why does this work ?!
pWpaKey->addKeyMgmt(std::string(*vArray));
}
else if(szKey == "Pairwise")
{
// ...
}
It looks like to work like a charm but I don't understant why! 看起来像是一种魅力,但我不明白为什么! vArray is Supposed to contain an adress right? vArray应该包含地址吗? And *vArray the "string" value. 和* vArray的“字符串”值。 So why when I "AND" an address with its value this give me an equality? 那么,为什么当我用其值“和”一个地址时却能给我带来平等的机会?
vArray && *vArray
is equivalent to (vArray != NULL) && (*vArray != NULL)
vArray && *vArray
等效于(vArray != NULL) && (*vArray != NULL)
It's first checking that the pointer vArray
isn't NULL
and, assuming it is not NULL
, checking that the pointer it points to isn't NULL
. 首先检查指针vArray
不为NULL
,并假设其不为NULL
,然后检查其指向的指针是否为NULL
。
The loop condition is 循环条件是
vArray && *vArray
This is basically shorthand for 这基本上是
(vArray != 0) && (*vArray != 0)
which is true if the char**
pointer is non-null and points to a char*
which is non-null. 如果char**
指针为非null并指向非null的char*
,则为true。
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