How to use execv() without warnings?

Question

I am working on MacOS-X Lion with GCC 4.2. This code works, but I get a warning I would like fix:

#include <unistd.h>
main()
{
    char *args[] = {"/bin/ls", "-r", "-t", "-l", (char *) 0 };
    execv("/bin/ls", args);
}

warning: deprecated conversion from string constant to 'char*'

I do not want the warning to be suppressed, I want not to have it at all. It is C++ code, not C.

Using a char *const (so exactly the type required by execv()) still produces the warning.

Thank you.

This give an error, not a warning. – Pietro Jun 09 '12 at 12:33 — Pietro, Jun 09 '12 at 12:33

score 4 · Accepted Answer · answered Jun 09 '12 at 14:45

4

This seems to be ok:

#include <unistd.h>
main()
{
    char const *args[] = {"/bin/ls", "-r", "-t", "-l", NULL };
    execv("/bin/ls", const_cast<char**>(args));
}

answered Jun 09 '12 at 14:45

Pietro

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score 1 · Answer 2 · answered May 28 '12 at 21:59

1

You are converting a string constant to a mutable character pointer, change using an implicit cast, the compiler is warning you that newer versions of the language will not allow this cast.

When defining a string litteral c++ understands it to mean a constant character array you have defined it as a mutable character array, change your code accordingly.

answered May 28 '12 at 21:59

awiebe

3,758
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awiebe: "You are converting a string constant". char *args[] is not constant. – Pietro Jun 09 '12 at 12:29
@Pietro: No but every single one of the literals you're trying to initialise it with _are_. – Lightness Races in Orbit Mar 25 '18 at 19:44

score 1 · Answer 3 · answered May 28 '12 at 23:16

1

Change:

char *args[] = {"/bin/ls", "-r", "-t", "-l", (char *) 0 };

To:

char *const args[] = {"/bin/ls", "-r", "-t", "-l", NULL };

answered May 28 '12 at 23:16

Jonathan

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I keep getting the same warning. – Pietro Jun 09 '12 at 12:21

score 1 · Answer 4 · answered Sep 05 '13 at 06:53

1

The only reason you get the warning is that you're using g++, not gcc. In pure C, you'd get absolutely no warnings. It's actually quite difficult to create warning-free C++ code from this. To be honest, I tried but did not succeed.

These hurdles are one of the reasons for the existence of a certain philosophical school. See more here.

answered Sep 05 '13 at 06:53

SzG

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After reading that comment by Linus Torvalds on C++, I would like to assist to a match between Torvalds and Stroustrup. I'm sure there would be a lot of blood :-) For what I am concerned, I would never go back from std::vector to malloc! – Pietro Sep 05 '13 at 14:16

score 1 · Answer 5 · answered Feb 15 '16 at 16:10

I do not know why the accepted answer was selected, it does not remove any warnings when I run this code....

I cannot confirm on your specific platform, but adding casts to each string constant has made the warnings go away for me.

#include <unistd.h>
main()
{
    char* const args[] = {(char*)"/bin/ls", (char*)"-r", (char*)"-t", (char*)"-l", (char*) 0 };
    execv("/bin/ls", args);
}

OR

#include <unistd.h>
main()
{
    char *args[] = {(char*)"/bin/ls", (char*)"-r", (char*)"-t", (char*)"-l", (char*) 0 };
    execv("/bin/ls", args);
}

It may be overly verbose and annoying, but the warnings go away.

I'm running this on: g++ (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4

The important difference is `char const*` vs `char * const`! First one is a pointer to *constant char*, second is a *constant pointer* to char. See also [Constness](https://en.cppreference.com/w/cpp/language/pointer#Constness) — Olaf Dietsche, Feb 24 '20 at 10:43

score 0 · Answer 6 · answered Apr 04 '13 at 21:55

0

change

char *args[] = {"/bin/ls", "-r", "-t", "-l", (char *) 0 };

to

char args[] = {(char*)"/bin/ls", (char*)"-r", (char*)"-t", (char*)"-l", 0 };

for usability u may will make a simple convert method.

answered Apr 04 '13 at 21:55

guest

1

1

You mean I should define an array of 8 bit chars an put 64 bit pointers inside? – Pietro Apr 05 '13 at 09:10

How to use execv() without warnings?

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