字符串到长数组在c++中

I understand that one can convert a string to a character array like so:

我知道可以将字符串转换为字符数组，如下所示:

string a = "abcdefgh";
char b[8];
strcpy(b, a.c_str());
cout << (int)b[3];

Here I get the output 100.

这里我得到输出100。

My questions is: How can I convert the string into an array of long. I am wondering how I can convert for example the string "abcdefgh" into an array long b[2]. The first long (b[0]) should be the long 0x61626364 and the second (b[1]) should be 0x65666768. If that makes sense. So

我的问题是:如何将字符串转换成一个长数组。我想知道如何将字符串“abcdefgh”转换为一个长b[2]的数组。第一个long (b[0])应该是long 0x61626364，第二个long (b[1])应该是0x65666768。如果这是有意义的。所以

cout << (unsigned int)b[0]

should output 1001633837924.

应该输出1001633837924。

2 个解决方案

#1

Try the following

试试以下

#include <iostream>
#include <iomanip>
#include <string>

int main() 
{
    std::string s( "abcdefgh" );
    long b[2] = {};

    for ( std::string::size_type i = 0, j = 0; i < 2 && j < s.size(); j++ )
    {
        b[i] = b[i] << 8 | ( unsigned char)s[j];
        if ( j % sizeof( long ) == sizeof( long ) - 1 ) i++;
    }

    std::cout << std::hex << b[0] << '\t' << b[1] << std::endl;

    return 0;
}

The output is

输出是

61626364    65666768

It would be even better to substitute statement

用代换语句会更好

        if ( j % sizeof( long ) == sizeof( long ) - 1 ) i++;

for

为

        if ( j % sizeof( *b ) == sizeof( *b ) - 1 ) i++;

In this case you could change the type of b without changing all other code. For example

在这种情况下，您可以更改b的类型，而无需更改所有其他代码。例如

#include <iostream>
#include <iomanip>
#include <string>

int main() 
{
    std::string s( "abcdefgh" );
    long long b[2] = {};

    for ( std::string::size_type i = 0, j = 0; i < 2 && j < s.size(); j++ )
    {
        b[i] = b[i] << 8 | ( unsigned char)s[j];
        if ( j % sizeof( *b ) == sizeof( *b ) - 1 ) i++;
    }

    std::cout << std::hex << b[0] << '\t' << b[1] << std::endl;

    return 0;
} 

The output is

输出是

6162636465666768    0

#2

You can use reinterpret_cast if your system uses the right endianess.

如果您的系统使用了正确的endianess，那么可以使用reinterpret_cast。

For example (it's not your expected output):

例如(它不是你的期望输出):

std::string a = "abcdefgh";
const long* b = reinterpret_cast<const long*>(a.data());
std::cout << std::hex << b[0] << " " << b[1] << std::endl;
// 64636261 68676665

If you want to get the other one, you'll have to code it yourself or use byte swap operations. Example with MSVC:

如果你想要得到另一个，你必须自己编码或者使用字节交换操作。与MSVC的例子:

#include<Bits.h>
// ....
std::cout << std::hex << _byteswap_ulong(b[0]) << " " << b[1] << std::endl;
// 61626364 68676665

It's easy to build the result with std::transform:

使用std::transform容易构建结果:

std::string a = "abcdefgh";
const long* b = reinterpret_cast<const long*>(a.data());
long c[2];
std::transform(b, b+2, c, _byteswap_ulong);
std::cout << std::hex << c[0] << " " << c[1] << std::endl; 
// 61626364 65666768