C语言 将 int 数组转换为 char 数组

Convert int array to char array

提问by CrazyCoder

I am having an array of integer say int example[5] = {1,2,3,4,5}. Now I want to convert them into character array using C, not C++. How can I do it?

我有一个整数数组 say int example[5] = {1,2,3,4,5}。现在我想使用 C 而不是 C++ 将它们转换为字符数组。我该怎么做？

采纳答案by JeremyP

Depending on what you really want, there are several possible answers to this question:

根据您真正想要什么，这个问题有几个可能的答案：

int example[5] = {1,2,3,4,5};
char output[5];
int i;

Straight copy giving ASCII control characters 1 - 5

直接复制给出 ASCII 控制字符 1 - 5

for (i = 0 ; i < 5 ; ++i)
{
    output[i] = example[i];
}

characters '1' - '5'

字符 '1' - '5'

for (i = 0 ; i < 5 ; ++i)
{
    output[i] = example[i] + '0';
}

strings representing 1 - 5.

代表 1 - 5 的字符串。

char stringBuffer[20]; // Needs to be more than big enough to hold all the digits of an int
char* outputStrings[5];

for (i = 0 ; i < 5 ; ++i)
{
    snprintf(stringBuffer, 20, "%d", example[i]);
    // check for overrun omitted
    outputStrings[i] = strdup(stringBuffer);
}

回答by karlphillip

#include <stdio.h>

int main(void)
{
    int i_array[5] = { 65, 66, 67, 68, 69 };
    char* c_array[5];

    int i = 0;
    for (i; i < 5; i++)
    {   
        //c[i] = itoa(array[i]);    /* Windows */

        /* Linux */
        // allocate a big enough char to store an int (which is 4bytes, depending on your platform)
        char c[sizeof(int)];    

        // copy int to char
        snprintf(c, sizeof(int), "%d", i_array[i]); //copy those 4bytes

        // allocate enough space on char* array to store this result
        c_array[i] = malloc(sizeof(c)); 
        strcpy(c_array[i], c); // copy to the array of results

        printf("c[%d] = %s\n", i, c_array[i]); //print it
    }   

    // loop again and release memory: free(c_array[i])

    return 0;
}

Outputs:

输出：

c[0] = 65
c[1] = 66
c[2] = 67
c[3] = 68
c[4] = 69

回答by unwind

You can convert a single digit-integer into the corresponding character using this expression:

您可以使用以下表达式将单个数字整数转换为相应的字符：

int  intDigit = 3;
char charDigit = '0' + intDigit;  /* Sets charDigit to the character '3'. */

Note that this is only valid, of course, for single digits. Extrapolating the above to work against arrays should be straight-forward.

请注意，这当然仅对个位数有效。将上述内容外推到数组上应该是直截了当的。

回答by peoro

You need to create the array, because sizeof(int)is (almost surely) different from sizeof(char)==1.

您需要创建数组，因为sizeof(int)（几乎肯定）不同于sizeof(char)==1.

Have a loop in which you do char_example[i] = example[i].

有一个你做的循环char_example[i] = example[i]。

---

---

If what you want is to convertan integer into a string you could just sum your integer to '0'but only if you're sure that your integer is between 0 and 9, otherwise you'll need to use some more sophisticated like sprintf.

如果您想要将整数转换为字符串，您可以将整数求和，'0'但前提是您确定整数在 0 到 9 之间，否则您需要使用更复杂的sprintf.

回答by Michael Smith

In pure C I would do it like this:

在纯 CI 中会这样做：

char** makeStrArr(const int* vals, const int nelems)
{
    char** strarr = (char**)malloc(sizeof(char*) * nelems);
    int i;
    char buf[128];

    for (i = 0; i < nelems; i++)
    {
        strarr[i] = (char*)malloc(sprintf(buf, "%d", vals[i]) + 1);
        strcpy(strarr[i], buf);
    }
    return strarr;
}

void freeStrArr(char** strarr, int nelems)
{
    int i = 0;
    for (i = 0; i < nelems; i++) {
        free(strarr[i]);
    }
    free(strarr);
}

void iarrtostrarrinc()
{
    int i_array[] = { 65, 66, 67, 68, 69 };
    char** strarr = makeStrArr(i_array, 5);
    int i;
    for (i = 0; i < 5; i++) {
        printf("%s\n", strarr[i]);
    }
    freeStrArr(strarr, 5);
}