在c数组的单个元素中输入字符串时，为什么以下代码给出显示的输出？

Question

#include <stdio.h>

int main(int argc, char *argv[])
{
    int i, n, m;

    scanf("%d %d", &n, &m);

    char s[m][n];

    for (i = 0; i < m; i++) {
        printf("the string --\n");
        scanf("%s", s[i]);
    }

    for (i = 0; i < m; i++) {
        printf("the strings are %s \n",s[i]);
        printf("\n");
    }

    return 0;
}

The output is:

2 2  
the string --
10  
the string -- 
11  

the strings are 1011 

the strings are 11

Why is the first string 1011 instead of 10 ?

Answer 1

In C, strings are represented as a sequence of char values, terminated by a null character ( 0 or '\\0' ). This means that to store a two-character string, you need space for three characters: the two characters of string content, plus the null terminator character.

Here, you've only allocated enough space for two characters in each string, but you need space for three.

So, it reads the first string into the array s[0] , but the null terminator doesn't fit, and so it overflows into the second array s[1] . Now your array of arrays s looks like this: {{'1', '0'}, {'\\0', ... }} .

Then, when it reads the second string into the array s[1] , it overwrites the overflowed null terminator from before. And the null terminator for the second string doesn't fit into its own array, so it overflows again into the rest of the stack. The program might crash here, or corrupt other data, because you're overflowing past the end of the array.

So now your array of arrays s ends up looking like this: {{'1', '0'}, {'1', '1'}} , followed by a '\\0' somewhere after the end of the array.

When printf goes to read your first string, it prints characters until it finds a null terminator. But it doesn't find one in the first string, so it keeps going, and hits the second string. It doesn't find one there either, and continues past the end of the array. In your case, luckily a null terminator was right there, but for all we know there could be something else.

To fix this, you need to allocate an extra character per string on line 9, for the null terminator:

    char s[m][n+1];

---

There's another problem here, however. What if your input gives you the wrong length? For example, what if your input says 2 3 , ie that the following strings will have a length of 3, but gives you the string foobar , which is 6 characters? Your code right now would overflow the buffer when it read that string, because it doesn't ensure it's the right length.

One way to avoid this would be to use gets_s instead of sscanf() for reading the strings on line 13:

        gets_s(s[i], n+1);

This will read at most n characters, so avoid crashing your program or creating a security issue. However, gets_s is a C11 function, so you may not be able to use it.

Answer 2

You must set column size to 3 if you insert 2 characters per string, 4 for 3 characters per string and so on. This because string in C have a termination character ('\\0') in the last position.

Answer 3

     #include <stdio.h>

int main(int argc, char *argv[])
{
    int i, n, m;

    scanf("%d %d", &n, &m);

    char s[m][n+1];

    for (i = 0; i < m; i++) {
        printf("the string --\n");
        scanf("%s", s[i]);
    }

    for (i = 0; i < m; i++) {
        printf("the strings are %s \n",s[i]);
        printf("\n");
    }

    return 0;
}

Answer 4

%s expects null terminated strings as an argument. When first string is read by scanf , there is not enough space for the null terminator within the allocated memory of first string. It will goes to the space next to the allocated space. Writing to unallocated space invokes undefined behavior.

While printing the strings with %s specifier, printf write the string character by character till it finds a null terminator '\\0' . Here it may be the case that both the strings 10 and 11 are stored one after another in memory, so printf writes the first string till it read the null character of second string.
Input n as 3 and you will get the desire results.

Answer 5

In my opninion using scanf to read strings is just pure evil. That said the array s[m][n] is just s[m*n] of course. That said that evil thing scanf is going to load on *s[0] 10\\n\\0 and on *s[1*n] or *s[2] 11 and *s will be 1011\\n\\0

And this is a monument to bad C coding. I guess it's just an example but if I was asked this question i would say: "Come on, get me real things"