如何在 GDB 调试器中打印字符串值而不是十六进制？

发布于 2024-10-29 21:10:50 字数 357 浏览 0 评论 0原文

(gdb) run hello
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /Users/doug/langs/c/test hello

Breakpoint 1, main (argc=2, argv=0xbffffa7c) at hw3b.c:14
14     if (argc != 2) {
(gdb) printf "%s", argv
??????(gdb)

我搜索了有关 SO 的其他问题，并尝试了找到的所有命令，但我不断收到？？？标记。这是为什么？

原文

(gdb) run hello
The program being debugged has been started already.
Start it from the beginning? (y or n) y
Starting program: /Users/doug/langs/c/test hello

Breakpoint 1, main (argc=2, argv=0xbffffa7c) at hw3b.c:14
14     if (argc != 2) {
(gdb) printf "%s", argv
??????(gdb)

I searched other questions on SO and I tried all the commands that I found but I keep getting ??? marks. Why is that?

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烟火散人牵绊 2024-11-05 21:10:50

argv 不是一个字符串，它是一个 char** - 指向可能多个 C 字符串中第一个的指针。

我认为您正在寻找：

print argv[0]
print argv[1]
...

或者如果您想使用 printf:

printf "%s\n", argv[0]

但在如此简单的情况下确实没有理由这样做，因为 gdb 确实知道如何打印 char* 字符串。

或者，如果您想变得更奇特，这是可行的：

print *argv@argc

语法 FOO@NUM 告诉它打印从 FOO 开始的 NUM 元素数组。我不知道为什么解除引用会起作用，但它确实起作用 - 我想 gdb 就是这样。有人启发我吗？

argv isn't a string, it's a char** - a pointer to the first of possibly multiple C strings.

I think you're looking for:

print argv[0]
print argv[1]
...

Or if you want to use printf:

printf "%s\n", argv[0]

But there's really no reason to in such a simple case, since gdb does know how to print char* strings.

Or, if you want to be fancy, this works:

print *argv@argc

The syntax FOO@NUM tells it to print an array of NUM elements starting at FOO. And I have no idea why the dereferencing works, but it does - I guess gdb is just nice like that. Someone enlighten me?

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