Question

我正在尝试将字符流转换为float *，但是不知何故无法获得正确的结果。

char  *char_data_  = static_cast<char *>("abcdefghijklmnopqrstuvwx");
float *float_data_ = reinterpret_cast<float *>(malloc(strlen(char_data_)/sizeof(float)));

printf("%ld\n", strlen(char_data_));

memcpy(float_data_, reinterpret_cast<float *>(char_data_), strlen(char_data_)/sizeof(float));

for ( auto n = 0 ; n < strlen(char_data_)/sizeof(float); n++) {
    printf("%f\n", *(float_data_ + n));
}

以下是我的结果，但其清除是错误的。有人可以看看吗？

16777999408082104352768.000000
0.000000
0.000000
0.000000
0.000000
0.000000

是否可以使用联合来解决此问题？

我希望每个字符流都有4个字节的块，并将其保存到浮点数组中。

Answer 1

如何使用union或memcpy将char *转换为float *？

使用memcpy和reinterpretcast：

#include <cstdio>
#include <string>
#include <cstring>
#include <cassert>
#include <cstddef>

void using_pointer(const char *s, size_t slen) {
    const float *f = reinterpret_cast<const float*>(s);
    for (size_t i = 0; i < slen/sizeof(float); ++i) {
        printf("%zu = %f\n", i, f[i]);
    }
}

void using_memcpy(const char* s, size_t slen) {
    float* f = new float[slen/sizeof(float)];
    memcpy(f, s, slen/sizeof(float)*sizeof(float));
    for (size_t i = 0; i < slen/sizeof(float); ++i) {
        printf("%zu = %f\n", i, f[i]);
    }
    delete f;
}

int main() {
    static_assert(sizeof(float) == 4, "");
    std::string stdstr(
        "\x00\x00\x80\x3f" // float 1.0
        "\x00\x00\x81\x3f" // just randomly changed 0x80 to 0x81 
        "\x00\x00\x82\x3f"
    , 4 * 3);

    printf("using_pointer:\n");
    using_pointer(stdstr.c_str(), stdstr.size());
    printf("using_memcpy:\n");
    using_memcpy(stdstr.c_str(), stdstr.size());

    std::string s2("abcdefghijklmnopqrstuvwx");
    printf("Last:\n");
    using_memcpy(s2.c_str(), s2.size());

    return 0;
}

将在http://www.onlinegdb.com使用的计算机上输出：

using_pointer:                                                                                                         
0 = 1.000000                                                                                                           
1 = 1.007812                                                                                                           
2 = 1.015625                                                                                                           
using_memcpy:                                                                                                          
0 = 1.000000                                                                                                           
1 = 1.007812                                                                                                           
2 = 1.015625                                                                                                           
Last:                                                                                                                  
0 = 16777999408082104352768.000000                                                                                     
1 = 4371022013021616997400576.000000                                                                                   
2 = 1138400301458999111806091264.000000                                                                                
3 = 296401655701622853703074578432.000000                                                                              
4 = 77151445562813935304650187079680.000000                                                                            
5 = 20076561220099179535696200212676608.000000

@edit：正如@drescherjm在评论中指出的那样，除非基于未定义的行为，否则在C ++中不可能使用并集在char和float表示形式之间进行转换。在C ++中，并集不用于更改字节的表示形式。联合用于最多存储多个对象之一。您可以一次将char数组或float数组存储在一个联合中，不能一次都存储，并且不能（至少不应该）使用C ++中的联合在float和char表示形式之间进行转换。

您的代码中发生了什么？

// static_cast<char*> from string is forbidden in iso C++. It's better to use  should use at least std::string.c_str()
// or use char char_data_[sizeof("abcdefghijklmnopqrstuvwx"]; memcpy(char_data_, "abcdefghijklmnopqrstuvwx", sizeof(char_data_));
char  *char_data_  = static_cast<char *>("abcdefghijklmnopqrstuvwx");
// you are allocating strlen("abcdefghijklmnopqrstuvwx")/4 = 24/4 = 6 bytes of memory. That's memory for 1 and a half float numbers.
float *float_data_ = reinterpret_cast<float *>(malloc(strlen(char_data_)/sizeof(float)));

// this will print '24\n'
printf("%ld\n", strlen(char_data_));

// you are copying 6 bytes of data from char_data_ to float_data_
// now float_data_ contains "abcdef" without '\0'
memcpy(float_data_, reinterpret_cast<float *>(char_data_), strlen(char_data_)/sizeof(float));

// this will print  16777999408082104352768.000000 on the first loop, which is "abcd" in hex in ascii
// then this invokes undefined behaviour cause of out of bound access
// you are trying to access elements number 2, 3, 4, 5 while float_data_ points to only 6 bytes, which is 1,5 float numbers, so even float_data_[1] is out of bound and undefined behaviour
for ( auto n = 0 ; n < strlen(char_data_)/sizeof(float); n++) {
    printf("%f\n", *(float_data_ + n));
}

使用union或memcpy将char 转换为float

1 个答案: