为什么在使用malloc（）动态分配内存后，字符数组打印出的字符数比需要少一个？

Question

The problem I am facing is that if I enter in 5 characters into an array using a loop that scanf's each character separately it only prints out 4 of the characters in the array (using a loop again). This problem does not arise if I do not dynamically allocate memory to the array.

 int main() {

    int nchars = 0;
    char* arr = NULL; 


    printf("Enter number of characters you want in string: "); scanf("%d", &nchars);

    arr = (char*)malloc(nchars * sizeof(char));

    printf("Enter characters you want in the array: \n");
    for (int i = 0; i < nchars; ++i) {
        scanf("%c\n", &arr[i]);
    }

    for (int i = 0; i < nchars; ++i) {
        printf("%c", arr[i]);
    }

    free(arr);
    return 0;
}

Terminal:

Enter number of characters you want in string: 5
Enter characters you want in the array:
a
m
i
i
i

amii (output)

If I do not put a \\n after the %c in the scan statement it only scans two chars and outputs them on new lines like this:

Enter number of characters you want in string: 5
Enter characters you want in the array:
a
m

a
m

I have tried different variations where I would create a temp array and strcpy() to the dynamically allocated char arr but still nothing. As well as:

arr = (char*)malloc((nchars + 1) * sizeof(char));

To see if allocating one more byte would add the last letter.

Answer 1

The problem is that scanf("%d", ...) leaves the newline character in the input stream. When you get to the scanf("%c\\n", ...) it reads the newline and the character you typed stays in the input stream, meaning you're off by one. You'll notice that in your output there's a blank line before amii is printed.

The solution is to have scanf skip whitespace. Change your loop to this:

for (int i = 0; i < nchars; ++i) {
    scanf(" %c", &arr[i]);
}

Putting a space before the %c means skip leading whitespace .