execvp function working without path

i am writing a running shell script with C programming. I have read about exec function although didn't understand much but I have read an example in which execvp is used like this

execvp(*argv, argv) ;

/* here argv is a char pointer array containing commands like ls -l

argv[0]-> ls argv[1]-> -l

*/

but it开发者_StackOverflow is used without giving file name as argument i idn't get it how it is working then . any one can please explain this as in the description it of execvp it is given to specified file name thanks so much

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In your case where you are actually passing Arg1: ls Arg2: ls -l After ensuring that your arguments are not NULL, this check is done

/* If it's an absolute or relative path name, it's easy. */
if (strchr(argv[0], '/')){
    execve(argv[0],argv,environ);
}
//In your case this would fail because argv[0] is not an absolute path.
//So now the search for argv[0] begins in the PATH
path = getenv("PATH")
//Now for each directory specified in path, it will attempt an execve call as
//For simplicity, I am calling each directoryname in PATH to be dir
execve(strcat(dir,argv[0]),argv,environ)
//If this generates an error called [ENOEXEC][1], then it's okay to continue searching in other directories, else quit searching and return the errorcode

I have provided a simplified and abstract view of execvp's working. You must look through the source code to understand the internal workings better