C ++中的矩形2D字符数组

Question

I'm using the 2D character arrays for storing multiple strings. The code below works fine & outputs the proper strings.

    char str[3][4];

    std::cin >> str[0] >> str[1] >> str[2] >> str[3];

    std::cout << str[0] << '\n' << str[1] << '\n' << str[2] << '\n' << str[3];

but when I try this

    char str[3][3];

    std::cin >> str[0] >> str[1] >> str[2];

    std::cout << str[0] << '\n' << str[1] << '\n' << str[2];

Then, if I input

abc

xyz

pqr

I get the output

abcxyzpqr

xyzpqr

pqr

What might be the possible explanation?

Answer 1

The size of your strings is 3 and you are giving 3 characters in input. Which means that it doesn't get the space to store the null character \\0 and therefore can't find the end of the string.

Answer 2

The reason is that C strings are null-terminated. This means that they are treated as ending when the first occurrence of a null character ('\\0', value 0) is encountered.

When you enter the three-character string "abc" into str[0], this will be represented in memory as {'a','b','c','\\0'}. Four characters, the last being the null-terminator.

This null terminator is then overwritten by the following input as str[0] is a 3-char string and thus, the null terminator will be written to str[0][3] which is equivalent to str[1][0].

Answer 3

It's because you're allocating 3 char s per string and they are of length 3 - so there's no \\0 at the end of the string. C++ strings should end with a \\0 to mark the end. When 4 char s are allocated, C++ automatically pads each string with a \\0 . It's safe to create arrays of at least n+1 char s when you need to store at most n char s in that.

It seems that the total 9 char s of str are contiguous, so the first string is of length 9. The cout statement goes to print from str[0][0] , and as there's no \\0 at the end of str[0] , it continues printing str[1] and str[2] .

Answer 4

These strings are named '0-terminated', because a literal string, in C, is a sequence of characters terminated by a 0 (null byte).

You are overflowing the terminator with the first character of the successive string.

Beware to undefined behaviour.