C ++不使用C标准库将字符串和整数转换为char *

Question

not using the C standard library do this in C++ ?

1. Convert string to char*
2. Covert int to char*

If i had to convert string to int in c++ use something like this

#include<iostream>
#include<string>
#include<sstream>
using namespace std;
int main()
{
    int num;
    string str="2020";
    (stringstream)"2020">>num;
    cout<<num+2;
    return 0;
}

Answer 1

"Convert string to char * " is impossible without the C++ standard library since string is a part of that library.

"Convert int to char * ": I assume you mean putting the decimal representation of an int in some buffer. This is how it can be done for unsigned ; doing the same for signed int means you have to take a possible - into account, and the corner case that arises from the fact that -INT_MIN is not well-defined.

unsigned n = SOME_VALUE;

char buffer[11]; // long enough for 32-bit UINT_MAX + NUL character
char *p = buffer + sizeof(buffer);

*--p = '\0';
do {
    *--p = '0' + n % 10;
    n /= 10;
} while (n);

p now points to the string representation of n .

Answer 2

I assume you mean you want a C++ solution using the C++ standard library (not using the C standard library). If so try the code below:

#include <iostream>
#include <sstream>

using namespace std;

int main()
{
    string stringString("2020");
    cout << "String String = " << stringString << endl;

    const char* charString = stringString.c_str();
    cout << "Char String = " << charString << endl;

    int charStringLen = stringString.size();
    for (int characterIndexCtr = 0; characterIndexCtr < charStringLen; ++characterIndexCtr)
    {
        cout << "Character At Index " << characterIndexCtr << " = " << charString[characterIndexCtr] << endl;
    }

    stringstream stringStream(stringString);
    int integerNumber;
    stringStream >> integerNumber;
    cout << "Integer = " << integerNumber << endl;
    cout << "Integer + 2 = " << integerNumber + 2 << endl;

    cout << "Press Enter To End Program ... ";
    cin.get();

    return 0;
}

Answer 3

I don't know whether you count Boost as stdlib, but lexical_cast can cast char*s to whatever you want:

char* foo = "123"
int bar = boost::lexical_cast(foo);

And the other way round:

int foo = 123;
std::string bar = boost::lexical_cast(foo);
your_function(bar.c_str());

It's using stringstream behind the scenes, but is a lot easier to use.
Also, you can't just convert an int to char*, as the memory the char* is pointing to has to be allocated somewhere.

Answer 4

#include <sstream>
#include <iostream>
#include <string>

int main (int argc, char* argv[])
{
    std::string str ("123");
    const char* c_str = str.c_str();

    char* so_bad = const_cast<char*>(c_str);

    std::stringstream ss;
    ss << so_bad;

    int int_value;
    ss >> int_value;

    std::cout << int_value;

    return 0;
}    

Answer 5

Convert string to char *

std::vector<char> vec( str.begin(), str.end() );
vec.push_back( '\0' );
char * data = &vec[0];

Convert string to int

std::istringstream iss(str);
int i;
if( !iss >> i )
{
    std::ostringstream oss;
    oss << "Invalid conversion from " << str << " to integer";
    throw std::invalid_argument( oss.str() );
}

Your second answer was close to the way to do it. Note there is a boost::lexical_cast which does pretty much the same, but has the huge downside of a meaningless bad_cast exception that gives no context information and therefore renders it almost useless in my opinion.