corrupted string

[@EE@]

Hi,
string s;

function(char* GetChArr);
function(&s[0])

what is the side effect of this?. s becomes nonassignable after return.

[2kaud]

Please provide full example code that compiles and shows the issue. What does function() do with GetChArr? If it gets chars and puts them into the memory pointed to by GetChArr and s is of type std::string, then you can only get .size(s) chars. You can't exceed this value. If the number of chars entered is less than s.size() then the remainder will not be changed. In all cases s.size() doesn't change.

[@EE@]

I have no access to the code for the function1. it is part of a library with no source.

when I pass:
char c[20];
function1(c);

then s = std::string(c),
then i can use the s inside a map to find a key.
if I use the s inside the map array to find a key it won't work. s inside the map become empty. In terms of making it work, it works that way but not sure what is happening.
sorry, if it is not clear.

[2kaud]

Depending upon what function() does and the size of s that is passed, this may have undefined behaviour. You may need to resize s first? What does function() do? As I mentioned above, if function() changes the contents passed then it cannot add more then s.size() chars.

In the code in post #3 function operates on a char array for up to 19 chars (plus the null terminator) which is fine (for a max of 19). This c-style char array is then assigned to a type std::string which again is fine.

As you can't change function() I suspect that the code in post #3 is how you'll need to use it. Or re-write the functionality of function() to work with a ref to std::string.

Does function() have a return value - or is it void? If function add chars to the given memory how does function() know how much memory has been allocated - as you don't seem to be passing a size to function()?

Without knowing what function() does or seeing it's code it seems that function() is ill-defined and is a prime candidate for a 'buffer overflow' memory error.