为什么在C中使用printf时字符串会被打印两次？

Question

My program asks the user to provide a string , which will be copied into an array of characters. Then, with a for loop, the program simply copies the elements of the first array into the second array.

int main() {

    int i;
    char string1[4], string2[4];

    // Get the first string
    printf("Insert your string: ");
    scanf("%s", string1);

    // Copy the values into the second array
    for (i = 0; i < 4; i++) {
        string2[i] = string1[i];
    }

    // Print the second string
    printf("%s", string2);
    return 0;
}

However, when I print the string using the printf() function the string gets printed twice.

Let's say I input the word

bars

The output will be

barsbars

Why is this happening?

Answer 1

char string1[4], string2[4];

4-element char array is not enough for 4-character strings. You need one more for the terminating '\\0' character.

Answer 2

Why?

TL;DR answer: undefined behaviour .

To explain, the problem here, with an input array defined like string1[4] , (4 elements only), an input string like bars will be overruning the allocated memory region (in attempt to store the terminating \\0 ), which in turn invokes undefined behaviour .

You should always take care of your input buffer length, like for an input array of string1[4] , your scanf() should look like

scanf("%3s", string1);

Answer 3

char bString []= {'s','t','r','i','n','g'};               
printf("bString:%s\n", bString);

Output:

bString:stringstring

Solution: Always include the terminating character

char bString []= {'s','t','r','i','n','g','\0'};    

Or simply write:

char bString [] = "string";