Using stl's vector.size() with printf

This is a discussion on Using stl's vector.size() with printf within the C++ Programming forums, part of the General Programming Boards category; Hi guys, I'm using this code: Code: vector<int> v; v.push_back(123); printf("size: %u\n", v.size()); And it retuns the result you would ...

  1. #1
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    Thumbs up Using stl's vector.size() with printf [solved]

    Hi guys,

    I'm using this code:
    Code:
    vector<int> v;
    v.push_back(123);
    printf("size: %u\n", v.size());
    And it retuns the result you would expect.
    But if you compile it you get this warning:
    Code:
    warning #181: argument is incompatible with corresponding format string conversion
                            printf("size: %u\n", v.size());
                                                 ^
    Is there any better solution than just writing that
    Code:
    (int) v.size()
    Can I use a different format string?
    Last edited by manuels; 08-18-2009 at 01:57 AM. Reason: marking it "solved"

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    C++ Witch laserlight's Avatar
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    That does not make sense: v.size() returns a std::vector<int>::size_type, which is an unsigned integer type. %u should be an acceptable format specifier.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
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    I know. But I just saw that the error only appears if I use Intel's icc.
    With g++ it works fine.

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    C++ Witch laserlight's Avatar
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    hmm... if it is feasible, consider using C++ I/O streams instead.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
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    yeah, would be a workaround. But my output line is quite complex so I would prefer a printf version.
    This warning annoys me since years and I want to know why it doesn't work like that.. in principle.
    Is it just an error by Intel's Compiler crew?

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    C++ Witch laserlight's Avatar
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    Quote Originally Posted by manuels
    But my output line is quite complex so I would prefer a printf version.
    You could consider using Boost.Format.

    Quote Originally Posted by manuels
    Is it just an error by Intel's Compiler crew?
    You could ask them, but they may be right: after all %u is meant to work with an unsigned int, but std::vector<int>::size_type is only specified as an unsigned integer type.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
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    I guess you could see if the compiler supports:
    Code:
    printf("size: %zu\n", v.size());
    But I don't think most C++ compilers support the %z modifier, and I'm not sure if vector<int>::size_type can be treated the same as a size_t.
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    Quote Originally Posted by laserlight View Post
    That does not make sense: v.size() returns a std::vector<int>::size_type, which is an unsigned integer type. %u should be an acceptable format specifier.
    unsigned integ[ral|er] type doesn't mean unsigned int type. so %u might be the wrong format specification.

    edit: hmm you said the correct way... not really sure what you think now

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    Quote Originally Posted by robwhit
    unsigned integ[ral|er] type doesn't mean unsigned int type. so %u might be the wrong format specification.
    Read post #6.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  10. #10
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    What if you try %ul instead? Maybe Intel thinks it's an unsigned long.
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    Quote Originally Posted by bithub View Post
    I guess you could see if the compiler supports:
    Code:
    printf("size: %zu\n", v.size());
    But I don't think most C++ compilers support the %z modifier, and I'm not sure if vector<int>::size_type can be treated the same as a size_t.
    Cool, %zu seems to work with Intel and GNU compiler.

    Thank you, folks!

  12. #12
    Super Moderator VirtualAce's Avatar
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    Size is actually a size_t type which can be easily cast to an unsigned int to remove the warning.

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