6

如何在char*不指定大小的情况下将无限字符读入变量?

例如,假设我想读取可能也需要多行的员工的地址。

4

3 回答 3

8

您必须首先“猜测”您期望的大小,然后使用malloc. 如果结果太小,您可以使用realloc将缓冲区调整为更大一点。示例代码:

char *buffer;
size_t num_read;
size_t buffer_size;

buffer_size = 100;
buffer = malloc(buffer_size);
num_read = 0;

while (!finished_reading()) {
    char c = getchar();
    if (num_read >= buffer_size) {
        char *new_buffer;

        buffer_size *= 2; // try a buffer that's twice as big as before
        new_buffer = realloc(buffer, buffer_size);
        if (new_buffer == NULL) {
            free(buffer);
            /* Abort - out of memory */
        }

        buffer = new_buffer;
    }
    buffer[num_read] = c;
    num_read++;
}

这只是我的想法,可能(阅读:可能)包含错误,但应该给你一个好主意。

于 2010-03-24T05:12:57.663 回答
1

只需要回答 Ivor Horton 第 3 版的《Beginning C》第 330 页的 Ex7.1。花了几个星期来锻炼。允许输入浮点数,而无需预先指定用户将输入多少个数字。将数字存储在动态数组中,然后打印出数字和平均值。在 Ubuntu 11.04 中使用 Code::Blocks。希望能帮助到你。

/*realloc_for_averaging_value_of_floats_fri14Sept2012_16:30  */

#include <stdio.h>
#include <stdlib.h>
#define TRUE 1

int main(int argc, char ** argv[])
{
    float input = 0;
    int count=0, n = 0;
    float *numbers = NULL;
    float *more_numbers;
    float sum = 0.0;

    while (TRUE)
    {
        do
        {
            printf("Enter an floating point value (0 to end): ");
            scanf("%f", &input);
            count++;
            more_numbers = (float*) realloc(numbers, count * sizeof(float));
            if ( more_numbers != NULL )
            {
                numbers = more_numbers;
                numbers[count - 1] = input;
            }
            else
            {
                free(numbers);
                puts("Error (re)allocating memory");
                exit(TRUE);
            }
        } while ( input != 0 );

        printf("Numbers entered: ");
        while( n < count )
        {
            printf("%f ", numbers[n]);  /* n is always less than count.*/
            n++;
        }
        /*need n++ otherwise loops forever*/
        n = 0;
        while( n < count )
        {
            sum += numbers[n];      /*Add numbers together*/
            n++;
        }
        /* Divide sum / count = average.*/
        printf("\n Average of floats = %f \n", sum / (count - 1));
    }
    return 0;
}

/* Success Fri Sept 14 13:29 . That was hard work.*/
/* Always looks simple when working.*/
/* Next step is to use a function to work out the average.*/
/*Anonymous on July 04, 2012*/
/* http://www.careercup.com/question?id=14193663 */
于 2012-09-14T15:56:04.213 回答
0

将一个 1KB 缓冲区(或 4KB)放在堆栈上,读取它直到找到地址的结尾,然后分配一个正确大小的缓冲区并将数据复制到它,怎么样?一旦你从函数返回,堆栈缓冲区就会消失,你只需要一次调用malloc.

于 2010-03-24T15:57:05.177 回答