连接器命令，退出code 1未能在第X code [英] Linker command failed with exit code 1 In Xcode

问题描述

我在X $ C $写C C程序，而我得到了这个问题：

I'm writing C program in Xcode, and I got this problem:

ld: 1 duplicate symbol for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

任何想法来解决呢？

Any idea to solve it?

注：我刚刚更新我的操作系统山狮昨天

Note: I've just update my OS to Mountain Lion yesterday.

推荐答案

一个常见的​​错误是的确定的在当你想要的声明一个头的象征的吧：

A common error is to define a symbol in a header when you wanted to declare it:

当你比如声明一个全局变量，而忘记了的extern 或你定义一个内联函数，而忘记了在线。在这种情况下，编译器在每次编译单元，包括这个头的象征和你结束了一个符号的多个定义。

When you for instance declare a global variable and forget the extern or you define an inline function and forget the inline. In these cases the compiler emits the symbol in each compile unit that includes this header and you end up with multiple definitions of a symbol.

反正你应该只是看问题的象征。

Anyway you should just look for the symbol in question.

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