Different output from sprintf function in c++

Question

The sprintf is not displaying the string message appropriately. The message to be displayed is Value out of range. Range is -2147483648 and 2147483647. However it is printed as Value out of range. Range is -2147483648 and 0.

#include <iostream>
#include <string>

int main()
{
    __int64 tmpminVal = -2147483648;
    __int64 tmpmaxVal = 2147483647;

    std::string strTemp = "Value out of range. Range is %d and %i ";
    char buffer[100];
    int n = sprintf (buffer, strTemp.c_str(), tmpminVal,tmpmaxVal);
    strTemp = buffer;
    std::cout << strTemp << std::endl;
    return 0;
}

Please provide the reason why it is does so.

%d and %i are for int but you provide __int64. This causes undefined behaviour. You will need to find out what your platform's specifier is for __int64. (If you use long long instead then %lld is the specifier). — M.M
– M.M, Commented Aug 5, 2014 at 11:37
%d and %i expect an int, not an __int64. What you see is probably the effect of tmpminVal being split into two chunks of 32 bits each. — Tobias Brandt
– Tobias Brandt, Commented Aug 5, 2014 at 11:37
thanks all. I defined the variables as long int and used the format specifier as %ld. and it is working fine. — suma
– suma, Commented Aug 5, 2014 at 12:17
@user657267, %I64d is not present in Linux. I wanted a general solution which works in both Windows and Linux. — suma
– suma, Commented Aug 5, 2014 at 12:18

ST3 · Accepted Answer · 2014-08-05 12:25:30Z

2

You can find printf parameters.

As you see %d is signed integer, integer in this case means 32-bit. Go further to specifiers table and you will see, that to print 64-bit (long long) you need to use specifier ll, so you need %lld, not %d.

Your result (-2147483648 and 0) is Undefined Behaviour

Also as I see from comments, you want cross platform solution, so you should use long long instead of __int64, as this one is Windows type.

edited Aug 5, 2014 at 12:25

answered Aug 5, 2014 at 11:43

ST3

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Comments

Sergej Christoforov · Accepted Answer · 2014-08-05 11:37:59Z

1

You passing two long long's to a function which expects two ints. Either change %d and %i to %ll or change __int64 to int.

answered Aug 5, 2014 at 11:37

Sergej Christoforov

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1 Comment

chux Over a year ago

Changing the format directive from "%d" to "%ll" is certainly wrong. Suspect you meant "%lld". Even "%lld" is not a great solution. "%lld" is for long long. long long may not be the same size as __int64. MS calls for "I64d" msdn.microsoft.com/en-us/library/aa261215(v=vs.60).aspx

Bill Lynch · Accepted Answer · 2014-08-05 12:43:48Z

So, as the other answers also say, you're invoking undefined behavior by passing a int64 to a format string expecting an int

Instead, you should use stdint.h and inttypes.h. On Linux, these will be included, and under Windows, you can include this project to use them: https://code.google.com/p/msinttypes/

An example use would be:

#include <iostream>
#include <string>
#include <inttypes.h>
#include <cstdint>

int main()
{
    int64_t tmpminVal = -2147483648;
    int64_t tmpmaxVal = 2147483647;

    std::string strTemp = "Value out of range. Range is %" PRId64 " and % " PRIi64 " ";
    char buffer[100];
    int n = sprintf (buffer, strTemp.c_str(), tmpminVal,tmpmaxVal);
    strTemp = buffer;
    std::cout << strTemp << std::endl;
    return 0;
}

You can also find some documentation about these header files at cppreference: http://en.cppreference.com/w/c/types/integer

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