CODEWITHSUNDEEP
c++gcccompiler-options
mmirzadeh
mmirzadeh

Reputation: 7079

What does the fpermissive flag do?

I'm just wondering what the -fpermissive flag does in the g++ compiler? I am getting:

error: taking address of temporary [-fpermissive]

which I can solve by giving the -fpermissive flag to the compiler.

EDIT: I just found what was causing the temporary address error part! I'm going to fix that part right now.

Upvotes: 160

Views: 265811

Answers (7)

Eugene W.
Eugene W.

Reputation: 446

As @cli_hlt mentioned

Bottom line: don't use it unless you know what you are doing!

It can do horrible things such that a compiler sometimes can cancel of constness of variables for std::map:

#include <map>
#include <vector>
#include <iostream>
#include <string>
struct B{
    std::map<std::string, int> m_map;
    std::vector<int> m_vector;
    B(){
        m_map["a"] = 1;
        m_map["b"] = 2;
        m_map["c"] = 3;
        m_vector.emplace_back(1);
        m_vector.emplace_back(2);
        m_vector.emplace_back(3);
    }
    const std::map<std::string, int>& getMap() const {
        return m_map;
    }
    const int& getMapValue(const std::string& key) const {
        return m_map.at(key);
    }
    const std::vector<int>& getVector() const {
        return m_vector;
    }
    const int& getVectorValue(const int& i) const {
        return m_vector[i];
    }
};

int main(){
    B b;
    auto& my_map = b.getMap(); // we get const ref here
    my_map["a"] = 10; // here we can modify it
    std::cout << "my_map[a]=" << my_map.at("a") << std::endl;

    auto& my_map2 = b.getMap();  // here we return already modified variable
    std::cout << "my_map2[a]=" << my_map2.at("a") << std::endl;

    auto& my_value = b.getMapValue("b");
    // my_value = 20; // compiler error
    // std::cout << "my_map[b]=" << my_value << std::endl;

    auto& my_vector = b.getVector();
    // my_vector[0] = 10; // compiler error
    // std::cout << "my_vector[0]=" << my_vector[0] << std::endl;

    const int a = 10;
    auto& a1 = a;
    // a1 = 100; // compiler error
}

As you can see you can't guarantee the constness of the map, however, the constness of a vector or value can be preserved. P.S. here I tested in the following compilers GCC 12.1, 9.1, 8.1, 7.1, 6.1. However, clang does not -fpermissive flag have this and it will catch the error.

Upvotes: 1

LarryM
LarryM

Reputation: 71

The general answer is that it "Downgrades some diagnostics about nonconformant code from errors to warnings."

Unfortunately, I haven't seen a specific list of things that it allows.

My main reason for posting an answer is to suggest that you avoid using it if at all possible. Instead, look at each error and see if it can be fixed. OP found and fixed what was causing their error. ("taking address of temporary" could be something like calling a function that returns a std::string object, and assigning something to the transient object's c_ptr() value.)

I was just reviewing a project that involved upgrading the version of gcc, and the developer added -fpermissive because there were suddenly a bunch of compilation errors. I noticed that one test was:

    if (myPointer == '\0')

I pointed out that it really should be:

    if (myPointer[0] == '\0')

The developer checked, and it turned out that every single thing flagged was a real error - some of which had been present for over 20 years.

Upvotes: 4

CarpenterSFO
CarpenterSFO

Reputation: 166

A common case for simply setting -fpermissive and not sweating it exists: the thoroughly-tested and working third-party library that won't compile on newer compiler versions without -fpermissive. These libraries exist, and are very likely not the application developer's problem to solve, nor in the developer's schedule budget to do it.

Set -fpermissive and move on in that case.

Upvotes: 5

David Schwartz
David Schwartz

Reputation: 182769

The -fpermissive flag causes the compiler to report some things that are actually errors (but are permitted by some compilers) as warnings, to permit code to compile even if it doesn't conform to the language rules. You really should fix the underlying problem. Post the smallest, compilable code sample that demonstrates the problem.

-fpermissive
Downgrade some diagnostics about nonconformant code from errors to warnings. Thus, using -fpermissive will allow some nonconforming code to compile.

Upvotes: 77

breakpoint
breakpoint

Reputation: 976

If you want a real-world use case for this, try compiling a very old version of X Windows-- say, either XFree86 or XOrg from aboout 2004, right around the split-- using a "modern" (cough) version of gcc, such as 4.9.3.

You'll notice the build CFLAGS specify both "-ansi" and "-pedantic". In theory, this means, "blow up if anything even slightly violates the language spec". In practice, the 3.x series of gcc didn't catch very much of that kind of stuff, and building it with 4.9.3 will leave a smoking hole in the ground unless you set CFLAGS and BOOTSTRAPCFLAGS to "-fpermissive".

Using that flag, most of those C files will actually build, leaving you free to move on to the version-dependent wreckage the lexer will generate. =]

Upvotes: 18

leftaroundabout
leftaroundabout

Reputation: 120711

When you've written something that isn't allowed by the language standard (and therefore can't really be well-defined behaviour, which is reason enough to not do it) but happens to map to some kind of executable if fed naïvely to the compiling engine, then -fpermissive will do just that instead of stopping with this error message. In some cases, the program will then behave exactly as you originally intended, but you definitely shouldn't rely on it unless you have some very special reason not to use some other solution.

Upvotes: 23

cli_hlt
cli_hlt

Reputation: 7164

Right from the docs:

-fpermissive
Downgrade some diagnostics about nonconformant code from errors to warnings. Thus, using -fpermissive will allow some nonconforming code to compile.

Bottom line: don't use it unless you know what you are doing!

Upvotes: 174

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