如何将 char 数组作为对函数和 memcpy 的引用传递
从早上起我就一直在努力解决这个问题。值 'pd' pm 在函数外部不会更改。有人能告诉我错误是什么吗？
void foo(u8 *pm, u8 *pd)
{
pm = "IWR ";
memcpy(pd, pm, sizeof(pm));
printf("in foo pm = %s, pd = %s \n", pm, pd);
}
int main()
{
u8 pm[5] = "0";
u8 pd[5] = "IWO ";
printf("pm = %s, pd = %s \n", pm, pd);
foo(pm, pd);
printf("after pm = %s, pd = %s \n", pm, pd);
}
调用 foo 后我的最终输出是 pm = (null) 和 pd = "IWO "。我认为“pm”也会改变价值。
（这是 ideone 上的代码，但在这种情况下 pm 打印为 0，而不是 (null)。）
pm = 0, pd = IWO
in foo pm = IWR , pd = IWR
after pm = 0, pd = IWR
I've been struggling to fix this since morning. The value 'pd' pm doesn't change outside the function. Could someone tell me what the mistake is?
void foo(u8 *pm, u8 *pd)
{
pm = "IWR ";
memcpy(pd, pm, sizeof(pm));
printf("in foo pm = %s, pd = %s \n", pm, pd);
}
int main()
{
u8 pm[5] = "0";
u8 pd[5] = "IWO ";
printf("pm = %s, pd = %s \n", pm, pd);
foo(pm, pd);
printf("after pm = %s, pd = %s \n", pm, pd);
}
My final output after call to foo is pm = (null) and pd = "IWO ". I thought that 'pm' also would change value.
(Here is the code on ideone, but in that case pm prints as 0, not (null). Why is this? )
pm = 0, pd = IWO
in foo pm = IWR , pd = IWR
after pm = 0, pd = IWR
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foo()中的sizeof(pm)是指针的大小。不是数组的大小，正如您所假设的那样。pm在函数之外不会改变，因为给定你的程序，pd肯定会改变。pm没有改变的原因是因为 C（以及您使用它的方式的 C++）是一种按值传递语言。 C 常见问题解答有一个关于您的问题的问题。sizeof(pm)in your functionfoo()is the size of a pointer. Not the size of the array, as it looks like you're assuming.pmdoesn't change outside the function, since given your program,pdmost certainly does. The reasonpmdoesn't change is because C (and C++ in the way you're using it) is a pass-by-value language. The C FAQ has a question about precisely your problem.您可以使用模板通过引用传递数组：
并以与之前使用
foo相同的方式使用它。请注意，这仅适用于数组，不适用于指针。请注意，
在原始函数中不执行任何操作（仅修改本地指针变量），并且在此修改后的函数中不起作用（无法分配给数组）。如果您想这样做，您还必须使用memcpy。
如果您不想使用模板，那么您必须将每个数组的大小传递到函数中（或使用哨兵值，但不要这样做），因为当您将数组传递到函数时（不通过引用）它会衰减为指针，虽然
sizeof(array)会给出数组中的字节数，但sizeof(pointer)只会给出字节数在一个不是你想要的指针中。You can pass an array by reference with a template:
And use it the same way you were using
foobefore. Note that this will only work with arrays, not with pointers.Note that
Doesn't do anything in your original function (just modifies the local pointer variable) and doesn't work in this modified one (can't assign to arrays). If you want to do that, you'll have to use
memcpyas well.If you don't want to use templates, then you'll have to pass the size of each array into the function (or use a sentinel value but don't) because when you pass an array into a function (without passing it by reference) it decays to a pointer, and while
sizeof(array)will give you the number of bytes in an array,sizeof(pointer)just gives you the number of bytes in a pointer which is not what you want.您好，通过执行 pm = "IWR" 您已经更改了 pm 指向 const 引用的内容。即，您已经完全更改了指针 pm （函数内部）以指向不同的位置。
谷歌改变链表的头部，你应该能够找到正确的解释
cslibrary.stanford.edu/103/LinkedListBasics.pdf [看看改变头指针]
Hi by doing pm = "IWR" you have changed what pm have been pointing to a const reference. i.,e you have completely changed the pointer pm ( inside the function ) to point to a different location.
Google for changing the head of linked list and you should be able to find proper explanation
cslibrary.stanford.edu/103/LinkedListBasics.pdf [ take a look changing head pointer ]