I have a string defined as std::string header = "00110033";
now I need the string to hold the byte v开发者_如何学Goalues of the digits as if its constructed like this
char data_bytes[] = { 0, 0, 1, 1, 0, 0, 3, 3};
std::string header = new std::string(data_bytes, 8).c_str());
I converted the initial string to int
array using atoi
. Now i'm not sure how to make the string out of it. Let me know if there is any better approach.
you could write a little function
string int_array_to_string(int int_array[], int size_of_array) {
string returnstring = "";
for (int temp = 0; temp < size_of_array; temp++)
returnstring += itoa(int_array[temp]);
return returnstring;
}
untested!
a slightly different approach
string int_array_to_string(int int_array[], int size_of_array) {
ostringstream oss("");
for (int temp = 0; temp < size_of_array; temp++)
oss << int_array[temp];
return oss.str();
}
Do this:
char data_bytes[] = { '0', '0', '1', '1', '0', '0', '3', '3', '\0'};
std::string header(data_bytes, 8);
Or maybe, you want to do this:
std::stringstream s;
s << data_bytes;
std::string header = s.str();
Demo at ideone : http://ideone.com/RzrYY
EDIT:
Last \0
in data_bytes is necessary. Also see this interesting output here: http://ideone.com/aYtlL
PS: I didn't know this before, thanks to Ashot I came to know this difference by experimenting!
char data_bytes[] = { 0, 0, 1, 1, 0, 0, 3, 3};
std::string str;
for(int i =0;i<sizeof(data_bytes);++i)
str.push_back('0'+data_bytes[i]);
Assuming you're using a "fairly normal" system where the numeric values of '0'
to '9'
are consecutive, you can just iterate over each element and subtract '0'
:
for(int i = 0; i < header.size(); ++i)
{
header[i] -= '0';
}
You can do this:
std::string header( data_bytes, data_bytes + sizeof( data_bytes ) );
std::transform( header.begin(), header.end(), header.begin(),
std::bind1st( std::plus< char >(), '0' ) );
If integ[] is the integer array, and s is the final string we wish to obtain,
string s="";
for(auto i=0;i<integ.size()-1; ++i)
s += to_string(ans[i]);
cout<<s<<endl;
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