CODEWITHSUNDEEP
cexec
baobobs
baobobs

Reputation: 703

execl not capturing all arguments

What could possibly be wrong with this execl statement? When I try to run it, the receiving executable complains that the argc is less than 3. When I print the argv contents, I get the following:

argv[1] = -1076146944
argv[2] = 0

Despite the arguments consisting of:

numJoeysStr = 6
randomNumSeedStr = 7

execl("/path/to/executable", "numJoeysStr", "randNumSeedStr", (char *)0);

FWIW, I tried NULL in replace of (char *)0). That didn't make a difference.

Chris Jester-Young resolved my biggest issue, but now I get the following after placing in the function twice:

argv[1] = -1075725068
argv[2] = -1075725056

I tried dereferencing, by doing:

printf("argv[1] = %d\n", *argv[1]);
printf("argv[2] = %d\n", *argv[2]);

Only to receive the following:

argv[1] = 110
argv[2] = 114

When I expected:

argv[1] = 6
argv[2] = 7

For the final problem, it turns out that passing the variables into execl without the quotes gave it the numbers I expected. I was somehow under the impression that all execl arguments (aside from the last) must be in quotes, even the variables.

Upvotes: 1

Views: 361

Answers (1)

C. K. Young
C. K. Young

Reputation: 223003

You actually need to specify "/path/to/executable" twice. The first one is the program to execute, and the second one is the argv[0] for the new process.

Upvotes: 4

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