如何通过一个常量数组常量到需要一个指针,而无需使用一个变量C / C ++函数? [英] How to pass a constant array literal to a function that takes a pointer without using a variable C/C++?
问题描述
如果我有一个原型,看起来像这样:
If I have a prototype that looks like this:
function(float,float,float,float)
我可以通过数值是这样的:
I can pass values like this:
function(1,2,3,4);
所以,如果我的原型是这样的:
So if my prototype is this:
function(float*);
有没有什么办法可以做到这样呢?
Is there any way I can achieve something like this?
function( {1,2,3,4} );
只是想找一个懒惰的方式做到这一点,而无需创建一个临时变量,但我似乎无法钉语法。
Just looking for a lazy way to do this without creating a temporary variable, but I can't seem to nail the syntax.
推荐答案
可以做到这一点在C99(不过的不的ANSI C(C90)或C ++任何当前的变体)具有的复合文字的。在血淋淋的细节,请参见C99标准的部分6.5.2.5。这里有一个例子:
You can do it in C99 (but not ANSI C (C90) or any current variant of C++) with compound literals. See section 6.5.2.5 of the C99 standard for the gory details. Here's an example:
// f is a static array of at least 4 floats
void foo(float f[static 4])
{
...
}
int main(void)
{
foo((float[4]){1.0f, 2.0f, 3.0f, 4.0f}); // OK
foo((float[5]){1.0f, 2.0f, 3.0f, 4.0f, 5.0f}); // also OK, fifth element is ignored
foo((float[3]){1.0f, 2.0f, 3.0f}); // error, although the GCC doesn't complain
return 0;
}
GCC还提供了这是一个扩展C90。如果用 -std = gnu90
编译(默认), -std = C99
或 -std = gnu99
,它会编译;如果你编译 -std = C90
,它不会。
GCC also provides this as an extension to C90. If you compile with -std=gnu90
(the default), -std=c99
, or -std=gnu99
, it will compile; if you compile with -std=c90
, it will not.
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