Question

我昨晚花了很多时间调试这段代码。我有两个数据文本文件，都包含18000个字符。我想将这些18000分成两个子串，每个子串100个字符，这样就可以进行180次迭代。

棘手的是，在前180次迭代中，两个子串的大小都很好。经过18次迭代后，子串的大小为0.

两个文件都已正确打开。我可以打印它们等等。我尝试以我能想到的所有可能方式分配子字符串，但到目前为止找不到解决方案。

int main(int argc, char const *argv[]) { 

    //Ive loaded two files into two strings buff1 and buff2 both size of 18000 chars
    //It works fine with small data example, I dunno why but eventually I have work with much more bigger data set
    //Id like to divide them into 100 char long pieces and do some stuff with that 

    char *substrA;      //substring for buff1
    char *substrB;      //substring for buff2

    substrA = malloc((wlen+1)*sizeof(char));        //word length wlen=100
    substrA = malloc((wlen+1)*sizeof(char));

    for (int i= 0; i <numOfSubProblems; ++i){   //numOfSubProblems = 18000/100

        strncpy(substrA, buff1+i*wlen, wlen);
        strncpy(substrB, buff2+i*wlen, wlen);
        substrA[wlen] = '\0';
        substrA[wlen] = '\0';

        int lenA = strlen(substrA);
        int lenB = strlen(substrB);
        printf("STRA a STR B: %d %d \n",lenA,lenB);

        DoSomething(substrA,substrB,i);  //some parser and other functionality
    }
    return 0;
}

Answer 1

strncpy不会终止目标字符串。所以你必须这样做

strncpy(subA, buff1+i*wlen, wlen);
subA[wlen] = '\0';
strncpy(subB, buff2+i*wlen, wlen);
subB[wlen] = '\0';

否则你不能使用strlen，并且在这样做的时候访问它们后面的缓冲区。

Answer 2

使用snprintf。

您可能没有处理格式化字符串，但至少它是一个理智的API。还要确保在确定子问题的数量时向上舍入：

#include <assert.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h>
#include <stdio.h>

#define PROBSIZE 18002

int main (int argc, char **argv) {
    char input[PROBSIZE];

    for (size_t i = 0; i < PROBSIZE; ++i) {
        input[i] = 'A' + (i % 10);
    }

    const size_t wlen = 10;
    char *subA = malloc (wlen + 1);
    if (!subA) {
        return EXIT_FAILURE;
    }

    for (int i = 0; i < (PROBSIZE + wlen - 1) / wlen; ++i) {
        /* If there's no error, guarantees `wlen` characters copied */
        int err = snprintf(subA, wlen + 1, "%s", input + i * wlen);

        if (err < 0) {
            fprintf(stderr, "snprintf encountered an error\n");
            return EXIT_FAILURE;
        }

        /* In absence of errors, we expect that the return value is
         * always >= wlen + 1, except the last iteration.
         */
        assert(err >= wlen + 1 || i == ((PROBSIZE + wlen - 1) / wlen) - 1);

        printf("%s\n", subA);
    }

    return EXIT_SUCCESS;
}

在C

2 个答案: