如何在 Xcode 中传递命令行参数
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How to Pass Command-Line Arguments in Xcode
提问by saulspatz
I'm trying to pass arguments to my Xcode C++ command-line tool. I think I'm following the help, and the answers I've found on Stack Overflow, but I'm not having any luck.
我正在尝试将参数传递给我的 Xcode C++ 命令行工具。我想我正在关注帮助，以及我在 Stack Overflow 上找到的答案，但我没有任何运气。
I'm using Xcode 7.2.1. I've edited the scheme to have two command-line arguments (an input file and an output file.) I've posted a screen shot of the editor below.
我正在使用 Xcode 7.2.1。我已经编辑了该方案，使其具有两个命令行参数（一个输入文件和一个输出文件）。我在下面发布了编辑器的屏幕截图。
When I run the code, it doesn't find the arguments. (I don't mean that it can't find the files on disk. The program aborts because argc is 1 instead of 3.)
当我运行代码时，它没有找到参数。（我不是说它找不到磁盘上的文件。程序中止是因为 argc 是 1 而不是 3。）
Can you tell me what I'm doing wrong?
你能告诉我我做错了什么吗？
EDIT: After playing around with it in response to the advice I received, it suddenly worked. I have no idea what changed. It works now with the dialog looking exactly like the screenshot I posted originally.
编辑：根据我收到的建议玩弄它后，它突然起作用了。我不知道发生了什么变化。它现在可以与我最初发布的屏幕截图完全一样的对话框使用。
回答by Logicrat
I'd suggest that you use only one argument set at a time. Each one of the checked lines represents a command line, not a single argument, so put both of your arguments on a single line, as in
我建议您一次只使用一个参数集。每一个选中的行代表一个命令行，而不是一个参数，所以把你的两个参数放在一行上，如
$PROJECT_DIR/detroit.txt $PROJECT_DIR/detroit.out
