所以我最近开始学习C语言，它开始变得比我想象的要复杂得多。我在解析字符串的函数上遇到了问题。我正在尝试做的是解析输入，例如：50 + 50或50+50，这样它就成为一个数组，其中50、+、50作为数组的不同元素。下面是parseString函数及其辅助函数：

int testnumber = 0;

void test() {
    testnumber++;
    printf("Test %d", testnumber);
    printf("\n");
}

bool isNumber(char *token) {
    int i;
    for (i = 0; token[i] != '\0'; i++) {
        if (!isdigit(token[i]))
            return false;
    }
    return true;
}

bool isOperator(char *token) {
    if (strlen(token) > 1)
        return false;
    switch (token[0]) {
      case '+':
      case '-':
      case '*':
      case '/':
      case '%':
        return true;
      default:
        return false;
    }
}

char **parseString(char *input) {
    test();
    int n_spaces = 0, i, start = 0, end;
    for (i = 0; i < strlen(input); i++) {
        if (input[i] == ' ') {
            n_spaces++;
        }
    }
    char **tokens = malloc(sizeof(char *) * (n_spaces + 1));
    for (i = 0; i < strlen(input); i++) {
        if (input[i] == ' ') {
            end = i;
            tokens = realloc(tokens, sizeof(char*) * (n_spaces + 1));
            tokens[n_spaces] = malloc(end - start + 1);
            strncpy(tokens[n_spaces], input + start, end - start);
            tokens[n_spaces][end - start] = '\0';
            start = i + 1;
            n_spaces++;
        }
    }
    end = strlen(input);
    tokens[n_spaces] = malloc(end - start + 1);
    strncpy(tokens[n_spaces], input + start, end - start);
    tokens[n_spaces][end - start] = '\0';
    n_spaces++;
    int j = 0;
    for (i = 0; i < n_spaces; i++) {
        char *temp = tokens[i];
        if (isNumber(temp)) {
            tokens[j++] = temp;
        } else
        if (isOperator(temp)) {
            tokens[j++] = temp;
        }
    }
    
    test();
    tokens[j] = NULL;
    return tokens;
}

下面也是我用来测试输出的主要函数：

int main() {
    // Declare variable 'size' and initialize it with 100
    size_t size = 100;

    // Allocate memory for a character array and assign it to pointer 'input'
    char *input = malloc(size);

    // Read input from standard input and store it in the character array 'input'
    fgets(input, (int) size, stdin);

    char **tokens = parseString(input);

    test();
    int i;
    for (i = 0; *(tokens + i); i++) {
        printf("%s\n", *(tokens + i));
    }

    test();
    // Free up memory.
    free(input);
    free(tokens);
    return 0;
}

我试着在main函数中添加一些printf，看看为什么它没有给予我期望的输出：

int main() {
    // Declare variable 'size' and initialize it with 100
    size_t size = 100;

    // Allocate memory for a character array and assign it to pointer 'input'
    char *input = malloc(size);

    // Read input from standard input and store it in the character array 'input'
    fgets(input, (int) size, stdin);

    char **tokens = parseString(input);

    printf("%s\n", tokens[0]);
    printf("%s\n", tokens[1]);
    printf("%s\n", tokens[2]);
    test();
    int i;
    /*for (i = 0; *(tokens + i); i++) {
        printf("%s\n", *(tokens + i));
    }*/

    test();
    // Free up memory.
    free(input);
    free(tokens);
    return 0;
}

但它只给出了输出：

Test 1
Test 2
50
+
(null)
Test 3
Test 4
Process finished with exit code 0

当我给予它输入50 + 50时，但当我给它输入50+50时，它给我输出：

Test 1
Test 2
(null)

Process finished with exit code -1073741819 (0xC0000005)