减去地址＆＃x2013;警告：＆＃x201c;在间接级别上有所不同＆＃x201d;

发布于 2025-02-10 09:42:01 字数 466 浏览 2 评论 0原文

为什么我会收到警告？ PCH已经是我获得的指针，当我想减去地址时，我使用＆amp; origi。

c4047' - '：'char*'间接水平与'char（*）[12]'
不同

// Substring

    char Origi[] = { "Hallo Welt." };
    char* pch = strstr(Origi, "Welt"); // “pch” is a pointer that in this example points 6 characters further than the start of “Origi”.
    printf("%d\n", pch - &Origi);
    printf("%c\n", Origi[pch - &Origi]);

原文

Why do I receive a warning? pch is already the pointer I got and when I want to subtract the addresses I use &Origi for that.

C4047 '-' : 'char*' differs in levels of indirection from 'char (*)[12]'

// Substring

    char Origi[] = { "Hallo Welt." };
    char* pch = strstr(Origi, "Welt"); // “pch” is a pointer that in this example points 6 characters further than the start of “Origi”.
    printf("%d\n", pch - &Origi);
    printf("%c\n", Origi[pch - &Origi]);

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你是我的挚爱i 2025-02-17 09:42:01

在摘要中：

printf("%d\n", pch - &Origi);

origi已经是类型的char*，因为当您将数组作为参数传递时，它会衰减到指向其第一个元素的指针时，传递其地址将使它成为一个指向一系列字符的指针，它与char的指针不同，并且会导致二进制操作中的类型不匹配。

为了使指针算术正常工作，操作数必须具有相同的类型。应该是：

printf("%d\n", pch - Origi);

                |_____|____
                          |
                          ---> same type -> char*

对于第二种情况，它的逻辑几乎是相同的，origi已经是指向char的指针。应该是：

printf("%c\n", Origi[pch - Origi]);

                       |_____|____
                                 |
                                 ---> same type -> char*

我承认MSVC发出的错误不是最好的，例如，它应该像GCC中一样，这更多的是：

错误：无效的操作数是二进制 - （拥有'char* '和'char （*）[12]'）

或更好的是，在Clang中：

错误：'char* '和'char （*）[12]'不是兼容类型的指示

In the snippet:

printf("%d\n", pch - &Origi);

Origi is already of type char* because when you pass an array as argument it decays to a pointer to its first element, passing its address will make it a pointer to array of chars, which is not the same as a pointer to char and will cause a type mismatch in the binary operation.

For the pointer arithmetic to work properly the operands must be of the same type. It should be:

printf("%d\n", pch - Origi);

                |_____|____
                          |
                          ---> same type -> char*

For the second case it's much the same logic, Origi is already a pointer to char. It should be:

printf("%c\n", Origi[pch - Origi]);

                       |_____|____
                                 |
                                 ---> same type -> char*

I'll admit the error issued by msvc is not the best, it should be something like in gcc, for example, which is much more on point: