c - c中指向字符数组的指针

Question

下面的程序对于值 9001 给出相同的输出。它应该为第一部分打印 9 ，为第二部分打印 1 。函数 words() 接受传递的正确值并根据需要返回适当的值。但在打印时出现问题，它不会打印函数传递的值。

感谢您的帮助。

#include<stdio.h>
#include<string.h>

char * words(int n);
const char * twenty[] = { "One ","Two ","Three ","Four ","Five ","Six 
","Seven ","Eight ","Nine ","Ten ","Eleven ","Twelve ",
"Thirteen ","Fourteen ","Fifteen ","Sixteen ","Seventeen ","Eighteen 
","Nineteen ","Twenty " };
const char * thirty[] = { "Thirty ","Fourty ","Fifty ","Sixty ","Seventy 
","Eighty ","Ninty " };

int main() {
    //get input from user to print no to words
    printf("Give no :");
    int x = 0;
    scanf("%d", &x);
    char * answer[] = { "","" };
    if (x < 0 || x > 999999) 
        printf("Give no between 1 and 999999\n");
    else {
        if (x / 1000 > 0) {
            answer[0] = words(x / 1000);
            answer[1] = words(x % 1000);
        }
        else 
            answer[0] = words(x % 1000);
        printf("Frist part is : %s\n", answer[0]);
        printf("Second part is : %s\n", answer[1]);
    }
    return 0;
}

char * words(int n) {
    char * ans = "";
    if (n == 0) 
        return ans;
    if (n / 100 > 0) 
        ans = strcat(twenty[n / 100 - 1], "Hundred ");
    if (n % 100 < 21 && n % 100 > 0) 
        ans = strcat(ans, twenty[n % 100 - 1]);
    else{
        if (n % 100>0) 
            ans = strcat(ans, thirty[(n % 100) / 10 - 3]);
        if (n % 10>0) 
            ans = strcat(ans, twenty[`enter code here`(n % 10) - 1]);
    }
    return ans;
};

Answer 1

在你的words()函数中你不能这样做，

strcat(ans, ....) 

或

strcat(twenty[n / 100 - 1], ...)

ans 和 twenty 中的元素没有足够的内存来容纳那么多字符。

其中一种方法是为 ans 分配缓冲区并使用它来返回单词。不要重写 twenty[] 中的字符串。

int max_ans_len = 1000;
char *ans = malloc(sizeof(char) * max_ans_len);
...

并且不要再次将strcat()的返回值赋给ans。还要确保您在其中写入的内容不超过 max_ans_len，并在使用完毕后将其释放。