What happens to dynamic allocated memory when calling execv()?

I'm writing a simple shell as an OS course assignment, I need to search in the PATH to find the program user typed in, once I find the right 开发者_如何学Cdirectory, I malloc a piece of memory just enough to hold the directory name plus the program name, and I pass it as the first argument to execv().

I could have statically allocated 100 characters or so, but having a limit makes me feel uncomfortable. So when execv() executes, is the heap cleaned up or is that piece of memory lost?

It's maybe not a lot of memory but I'm just curious.

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When you exec(), the entire process is (a) ended, so all resources including dynamic memory and some fd's as below, are reclaimed by the operating system, and (b) replaced: code, data, threads, ...

Re file descriptors, from "man execve":

File descriptors open in the calling process image remain open in the new process image, except for those for which the close-on-exec flag is set (see close(2) and fcntl(2)). Descriptors that remain open are unaffected by execve().