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gcc compiler didn't error on function with no return statement

+1 vote
75 views

I need some help in understanding why my GCC didn't consider this an issue. I have a function that was constructing a path to a daemon program based on the location of the shared object file where this code is. Something similar to this:

namespace {
 std::string ConstructPath()
 {
 int lastSlash(0);
 std::string pathVar;
 Dl_info dl_info;
 memset(

 if((dladdr((void*)ConstructPath, 
 }

 pathVar = dl_info.dli_fname;
 lastSlash = pathVar.find_last_of('/');
 if(std::string::npos == lastSlash)
 {
 // no slashes given ... must be that *.so
 // is in the current directory
 pathVar = "mydaemond";
 }
 else
 {
 pathVar.erase(pathVar.begin() + (lastSlash + 1), pathVar.end());
 pathVar.append("mydaemond");
 }

 // first check if we can find the daemon
 {
 // introducing sub-scope to ensure the file object is closed
 std::ifstream test(pathVar.c_str());
 if(!test.good())
 {
 throw std::runtime_error("cannot find mydaemond");
 }
 }

 // *** the below statement wasn't there originally, the
 // *** function simply exited after the forced-scope block above,
 // *** however, the function *did* have the return type
 return pathVar;
 }
}

My comments above the final return statement illustrate what my question is about. Why wasn't this a problem? There was no return statement and yet, the code compiled fine. I'm using GCC 4.4.4 on CentOS 6.2. Is this just a problem with the 4.4.4 compiler that was fixed? I'm betting there's some subtlety in C++ here that I'm not yet aware of and I'd like to be schooled.

posted Jun 20, 2013 by anonymous

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1 Answer

+1 vote
 
Best answer

Failing to provide a return value is legal C, but undefined behaviour. So gcc has to accept it.

You only get warnings if you enable them. Normally, you will want your code to be warning-free when compiled with "-Wall". Many people use "-Wextra" as well - I personally enable many of the other warning flags as well.

answer Jun 20, 2013 by anonymous
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