I am attempting to invoke a command with execvp/execlp. If a file appears in my PATH before the executable desired and has the same name as the executable, the first occurrence of the file name is used as the executable to invoke. Even though the first file is NOT marked as executable.
The attached testcase illustrates this.
Here's a simple shell log:
bash-3.00$ gcc test2.c -o test2
bash-3.00$ PATH=/usr/bin:.
bash-3.00$ test2 nopathnoext
NoPathNoExt
bash-3.00$ touch echo
bash-3.00$ ls -l echo
-rw-r--r-- 1 test None 0 Jan 21 15:39 echo
bash-3.00$ test2 nopathnoext
NoPathNoExt
bash-3.00$ PATH=.:/usr/bin
bash-3.00$ test2 nopathnoext
nopathnoext: No such device or address
bash-3.00$ rm -f echo
bash-3.00$ test2 nopathnoext
NoPathNoExt
bash-3.00$ echo garbage > echo
bash-3.00$ test2 nopathnoext
nopathnoext: Permission denied
bash-3.00$ ls -l echo
-rw-r--r-- 1 test None 8 Jan 21 15:40 echo
bash-3.00$ test2 nopathext
NoPathExt
bash-3.00$ test2 path
Path
Is this normal behavior for execlp/execvp?
Shouldn't the execution permission be set in order to
execute it?
Thanks for your help,
Martin
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cygcheck.out
Description: 3924260979-cygcheck.out
test2.c
Description: 2263429771-test2.c
test2.sh
Description: 809292912-test2.sh
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