被++一样+ = 1的指针？ [英] Is ++ the same as += 1 for pointers?

问题描述

我想重构我的一些旧的C code，我很好奇，如果我可以取代所有 PTR ++ 与 PTR + = 1 ，其中 PTR 是一些指针，不改变任何行为。下面是我的意思，从K＆放一个例子; R第5.3节：

I'd like to refactor some old C code of mine, and I was curious if I can replace all ptr++ with ptr += 1 where ptris some pointer, without changing any behavior. Here's an example of what I mean, from K&R Section 5.3:

/* strlen: return length of string s*/
int strlen(char *s)
{
    int n;
    for (n = 0; *s != '\0'; s++)
        n++;
    return n;
}

当我与 S + = 1 替换取值++ ，我得到相同的结果，但我想知道这将是对所有类型的情况。我还做了一个测试为 INT 取值：

When I replace the s++ with s += 1, I get the same results, but I'm wondering if this will be the case for all types. I also did a test for ints:

int size = 10;
int *int_array = (int*)calloc(size, sizeof(int));
for (int i = 0; i < size; i++)
    int_array[i] = i;

for (int i = 0; i < size; i++) {
    printf("*int_array = %d\n", i, *int_array);
    int_array++;
}

如果我更换行 INT_ARRAY ++; 与 INT_ARRAY + = 1; ，我得到相同的结果

If I replace the line int_array++; with int_array += 1;, I get the same result.

在思考这个多一些之后，我意识到，如果该值在一个前pression使用有可能是一个问题。难道是我安全刚搬到增量到像这样另一行：

After thinking about this some more, I realize there could be a problem if the value is used in an expression. Would it be safer I just moved the increment onto another line like so:

int a = 5;
int b = a++;

将变成：

int a = 5;
int b = a;
a += 1;

结论

我认为可能是一个问题，递增的不同类型的指针，是没有问题的。见@ bdonlan的为什么的原因响应。

What I thought could be a problem, incrementing pointers of different types, is not a problem. See @bdonlan's response for the reason why.

这并不意味着你可以替换所有 X ++ 与 X + = 1 ，并期望相同的行为。你可以，但是，替换 ++ X 与（X + = 1）安全，因为它们是等价的。

This doesn't mean that you can replace all x++ with x += 1 and expect the same behavior. You can, however, replace ++x with (x += 1) safely, since they are equivalent.

推荐答案

A + = 1 等同于 ++中的（C99§6.5.3.1/ 2）。在像 INT B A线= A ++; 这意味着它的不的等同于 A ++ ; A ++ 将返回的 A ，而老的价值+ = 1 返回新的值。

a += 1 is equivalent to ++a (C99 §6.5.3.1/2). In a line like int b = a++; this means it is not equivalent to a++; a++ would return the old value of a, while a += 1 returns the new value.

请注意，如果你不使用 A ++的结果（也就是说，你有一个包含一份声明中只是 A ++; ），那么它们实际上是相同的。

Note that if you don't use the result of a++ (ie, you have a statement containing just a++;), then they are effectively identical.

另外，还要注意_all指针运算中所指向的类型的大小（§6.5.6/ 8）为单位进行。这意味着：

Also, note that _all pointer arithmetic is done in increments of the pointed-to type's size (§6.5.6/8). This means that:

ptr = ptr + x;

相当于：

ptr = (ptr_type *)( (char *)ptr + x * sizeof(*ptr) );

这是你是否使用相同的 + ， ++ ， + = 或 [] （ p [X] 完全等同于 *（p + X）;你甚至可以做这样的事情 4你好] ）

This is the same whether you use +, ++, +=, or [] (p[x] is exactly equivalent to *(p + x); you can even do things like 4["Hello"] because of this).

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