使用垃圾值打印，c

Question

我遇到了问题，我无法弄明白为什么。在主要我创建2个字符串数组。该函数接受2个字符串并根据以下要求创建适当大小的新数组：例如：数组1：aviv，数组2：12，新数组：a12v12i12v新数组必须完全相同！然后将新阵列发送到main和main以进行打印。我也打印垃圾值。我检查了新数组的大小，它的大小合适。我的代码：

 char* CreateString(char* str1, char* str2)
    {
        int length1, length2, length3 = 0;
        int i,j, index_help = 0;
        char *str3 = NULL;
        length1 = strlen(str1);
        length2 = strlen(str2);
        for (i = 0; i < length1; i++) // Check the size of the new array

        {
            length3++;
            if (i == (length1 - 1))
            {
                break;
            }
            for (j = 0; j < length2; j++)
            {
                length3++;
            }
        }
        str3 = (char*)malloc(length3+1  * sizeof(char));
        if (str3 == NULL)
        {
            printf("There is not enough memory space\n");
            return 0;
        }
        for (i = 0; i < length1; i++) //Copying data

        {

            str3[index_help] = str1[i];
            if (i == (length1 - 1))
            {
                break;
            }
            for (j = 0; j < length2; j++) 
            {
                index_help++;
                str3[index_help] = str2[j];
            }
            index_help++;

        }
        return str3;
    }

    int main()
    {
        char *str1 = NULL, *str2 = NULL,*str4=NULL;
        int size1, size2,i;
        printf("enter the size of string number 1:\n");
        scanf("%d", &size1);
        printf("enter the size of string number2 :\n");
        scanf("%d", &size2);
        str1 = (char*)malloc((size1 + 1) * sizeof(char));
        if (str1 == NULL)
        {
            printf("There is not enough memory space\n");
            return 0;
        }
        str2 = (char*)malloc((size2 + 1) * sizeof(char));
        if (str2 == NULL)
        {
            printf("There is not enough memory space\n");
            return 0;
        }
        printf("Enter a value for the first string (the size is:%d):\n", size1);
        scanf("%s", str1);
        printf("Enter a value for the second string (the size is:%d):\n", size2);
        scanf("%s", str2);
        str4 = CreateString(str1, str2);
printf("%s",str4);
        printf("\n");
        return 0;
    }

Answer 1

该函数应该声明为

 char* CreateString( const char* str1, const char* str2);

因为在函数中没有更改字符串str1和字符串str2。

您必须使用终止零附加结果字符串。

在不使用循环的情况下，可以更简单地计算结果字符串的长度。

由于函数strlen具有返回类型size_t，因此接受函数调用结果的变量也应该声明为具有类型size_t而不是类型int。

该功能可以按照以下方式实施，如示范程序中所示。

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

char * CreateString(const char *s1, const char *s2)
{
    size_t n1 = strlen(s1);
    size_t n2 = strlen(s2);

    size_t n3 = n1 == 0 ? 0 : n1 + (n1 - 1) * n2;

    char *s3 = ( char * )malloc(n3 + 1);

    if (s3)
    {
        if (n2 == 0)
        {
            strcpy(s3, s1);
        }
        else
        {
            char *p = s3;

            while (*s1)
            {
                *p++ = *s1++;
                if (*s1)
                {
                    strcpy(p, s2);
                    p += n2;
                }
            }

            *p = '\0';
        }
    }

    return s3;
}

int main( void )
{
    char *p = CreateString("aviv", "12");

    printf("\"%s\"\n", p);

    free(p);

    p = CreateString("", "12");

    printf("\"%s\"\n", p);

    free(p);

    p = CreateString("a", "12");

    printf("\"%s\"\n", p);

    free(p);

    p = CreateString("av", "12");

    printf("\"%s\"\n", p);

    free(p);

    p = CreateString("av", "");

    printf("\"%s\"\n", p);

    free(p);

    return 0;
}

程序输出是

"a12v12i12v"
""
"a"
"a12v"
"av"