CODEWITHSUNDEEP
cshellexec
Mehdi
Mehdi

Reputation: 2228

Execute shell program with execv

I'm writing in C langage a program that contain these lines:

void main(int argc, char *argv[]) {
    char* file=argv[1];
    char* arguments[] = { "sh", argv[2], argv[3], argv[4], file, NULL };
    execv("/bin/sh", arguments);  
}

The file is prog.sh which contain a simple sum of arguments:

expr $1 + $2 + $3

When I run the program by ./main prog.sh 1 2 3 I obtain an error which is

/bin/sh: 0: Can't open 1

While I expect the output 6 (sum of 1 2 3)

Upvotes: 0

Views: 9096

Answers (1)

that other guy
that other guy

Reputation: 123470

Look at your arguments:

char* arguments[] = { "sh", argv[2],argv[3],argv[4],file, NULL };

When you run ./main prog.sh 1 2 3, you end up calling:

sh 1 2 3 prog.sh

You should instead make the script the first argument:

char* arguments[] = { "sh", file, argv[2],argv[3],argv[4], NULL };

thereby calling

sh prog.sh 1 2 3

Upvotes: 2

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