ShellExecute() file in same directory

[nbgoku]

1.exe has the following code

ShellExecute(0, "open", "2.exe", NULL, NULL, SW_SHOWNORMAL);

1.exe and 2.exe are in the same directory

now this works cool only if i give the full path to 2.exe, but the thing is, the 2.exe is in the same directory as the .exe file im launching it from, so how do i code ShellExecute() to open 2.exe which is in the same directory without having to give the full path

 

[Cagliostro]

I guess something like:
".\\2.exe"

Ion Filipski

 

[ArkM]

Obtain caller's module path in 1.exe and build full path of 2.exe for ShellExecute():

char path[_MAX_PATH];
GetModuleFileName(0,path,sizeof path); // curr module path
char* p = strrchr(path,'\\'); // where exe name pos...
strcpy(p+1,"2.exe"); // Now pass path to ShellExecute...

Be careful: No errors checking in my snippet...
It seems using .\ is not robust method because of no guarntee that current dir is the same as module (1.exe) dir. It seems it is the main source of your problem...

 

[Zyrenthian]

If you have the files in your DEBUG/RELEASE directory and you execute the program from dev studio, you may see it looking at the project folder and NOT the debug/release directories. I have noticed this when running in debug.

Matt

 

[nbgoku]

sweet thanks yall!


im curious though, has does a person go about finding information like using .\\ to mean the same directory?