Question

我试图编写代码来附加两个没有函数strcat的字符串，但它不起作用。它仅在输入为单个char时有效。

以下是代码：

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

/*
    Return the result of appending the characters in s2 to s1.
    Assumption: enough space has been allocated for s1 to store the extra
    characters.
*/
char* append (char s1[], char s2[]) {
    int s1len = strlen (s1);
    int s2len = strlen (s2);
    int k;
    for (k=0; k<s2len; k++) {
        s1[k+s1len] = s2[k];
    }
    return s1;
}

int main ( ) {
    char str1[10];
    char str2[10];
    while (1) {
        printf ("str1 = ");
        if (!gets(str1)) {
            return 0;
        };
        printf ("str2 = ");
        if (!gets(str2)) {
            return 0;
        };
        printf ("The result of appending str2 to str1 is %s.\n",
            append (str1, str2));
    }
    return 0;
}

我该如何解决？

Answer 1

您的代码中存在一些错误：

请勿使用gets 。这个功能是危险的，因为它没有采取考虑缓冲区的大小。请改用fgets。

您的append函数未写入'\0' - 终止字节。这应该是

char* append (char s1[], char s2[]) {
    int s1len = strlen (s1);
    int s2len = strlen (s2);
    int k;
    for (k=0; k<s2len; k++) {
        s1[k+s1len] = s2[k];
    }
    s1[k+s1len] = 0; // writing the 0-terminating byte
    return s1;
}

str1可能会因为持有两个字符串而缩短。如果str2包含"Hello" 并且str2包含"World!"，您将溢出缓冲区。制作缓冲区更大。

如果您自己编写strcat，我会传递目的地的大小缓冲区也是如此，这样你就不会溢出缓冲区：

char *mystrcat(char *t1, const char *t2, size_t maxsize)
{
    if(t1 == NULL || t2 == NULL)
        return NULL;

    size_t s1 = strlen(t1);
    size_t s2 = strlen(t2);

    size_t i;
    for(i = 0; i < s2 && (s1 + i) < maxsize - 1 ; ++i)
        t1[i + s1] = t2[i];

    t1[i + s1] = 0; // terminating the

    return t1;
}


int main(void)
{
    char str1[30] = "Hello ";
    char str2[30] = "World!";

    printf("mystrcat(\"%s\", \"%s\") = %s\n", str1, str2,
            mystrcat(str1, str2, sizeof str1));

    char str3[100] = "This is a long sentence";

    printf("mystrcat(\"%s\", \"%s\") = %s\n", str1, str3,
            mystrcat(str1, str3, sizeof str1));

    char line[100];
    printf("Enter some text: ");
    fflush(stdout);

    fgets(line, sizeof line, stdin);
    line[strcspn(line, "\n")] = 0; // removing possible newline

    strcpy(str3, "User input was: ");

    printf("mystrcat: %s\n", mystrcat(str3, line, sizeof str3));

    return 0;
}

那将返回

mystrcat("Hello World!", "World!") = Hello World!
mystrcat("Hello World!This is a long se", "This is a long sentence") = Hello World!This is a long se
Enter some text: ABC DEF user input is great
mystrcat: User input was: ABC DEF user input is great

Answer 2

您的结果字符串未正确终止。

C中的字符串始终以NUL（0）字符终止。
gets不安全，请改用fgets。
养成检查缓冲区大小的习惯。

为了给你一个想法，这里是一个使用fgets +缓冲区大小检查的简约实现：

#include <stdio.h>  // fprintf, fgets
#include <string.h> // strlen

const char *concatenate(char *dst, size_t sz, const char *s1, const char *s2) {
    size_t l1 = strlen(s1);
    size_t l2 = strlen(s2);

    // Check for overflow
    if ((l1 + l2 + 1) > sz) {
        return NULL;
    }

    // Copy first string
    for (size_t i = 0; i < l1; ++i) {
        dst[i] = s1[i];
    }

    // Copy second string
    for (size_t i = 0; i < l2; ++i) {
        dst[l1 + i] = s2[i];
    }

    // Add NUL terminator
    dst[l1 + l2 + 1] = 0;

    return dst;
}

int main() {
    // Allocate two strings (9 chars max.)
    char first_string[10];
    char second_string[10];
    char concatenated_string[20];

    // Read first string from stdin
    fprintf(stdout, "str1 = ");
    // !!! fgets return value check omitted for simplicity.
    fgets(first_string, sizeof(first_string), stdin);

    // Read second string from stdin
    fprintf(stdout, "str2 = ");
    // !!! fgets return value check omitted for simplicity.
    fgets(second_string, sizeof(second_string), stdin);

    const char *tmp = concatenate(concatenated_string, sizeof(concatenated_string), first_string, second_string);

    if (!tmp) {
         fprintf(stderr, "would overflow\n");
    } else {
         fprintf(stdout, "concatenated: %s\n", concatenated_string);
    }

    return 0;
}

附加两个字符串

2 个答案: