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Why does bash seem to parse `sh -c` commands differently when called via `execl`?

When I do this in a shell:
```plain
$ /bin/sh -c 'echo hello world'
hello world
```
it's my understanding that I'm running a process with `argv = {"/bin/sh", "-c", "echo hello world"}` That is, the quotes cause all three words to be passed as one argument to `sh`, and that instance of `sh` is responsible for splitting it on spaces before executing it.

I'm trying to have a C program do the same thing. I tried this:
```c
#include <unistd.h>

int main() {
	execl("/bin/sh", "-c", "echo hello world", NULL);
}
```

but for some reason the shell seems to parse its arguments differently when called this way:

```plain
$ ./a.out 
-c: echo hello world: No such file or directory
```

That is, it's getting `argv[2] = "echo hello world"`, and rather than parsing that as a shell command, deciding that must all be the `argv[0]` of the subprocess it's been asked to start.

My `/bin/sh` is `bash`. Using `/bin/bash` explicitly, or `/bin/zsh`, does the same thing, but busybox `sh` seems to do what I expect:
```c
$ cat /tmp/what.c 
#include <unistd.h>

int main() {
	execl("/bin/busybox", "sh", "-c", "echo hello world", NULL);
}

$ gcc /tmp/what.c && ./a.out 
hello world
```

Why (and for that matter, _how_) do `bash` and `zsh` behave this way? How can I get them to not do that?