CODEWITHSUNDEEP
c++arraysstringinitializationstring-literals
chasep255
chasep255

Reputation: 12175

Does the null character get automatically inserted?

Is the C++ compiler supposed to automatically insert the null character after the end of the char array? The following prints "Word". So does the C++ compiler automatically insert a null character at the last position in the array?

int main()
{
    char x[] = {'W', 'o', 'r', 'd'};
    cout << x << endl;
    return 0;
}

Upvotes: 1

Views: 3219

Answers (8)

xitong
xitong

Reputation: 349

Although it is not guaranteed, it seems my implementation always adds a \0 even if I am not requested.

I tested noNull[2] is always \0

char noNull[] = {'H', 'i'};
for (int i = 2; i < 5; ++i) {
    if (noNull[i] == '\0') {
        std::cout << "index " << i << " is null" << std::endl;
    } else {
        std::cout << "index " << i << " is not null" << std::endl;
    }
}

output

index 2 is null
index 3 is not null
index 4 is not null

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 310930

In C/C++ string literals are automatically appended with terminating zero. So if you will write for example

char x[] = "Word";

then the size of the array will be equal to 5 and the last element of the array will contain '\0'.

When the following record is used as you showed

char x[] = {'W', 'o', 'r', 'd'};

then the compiler allocates exactly the same number of elements as the number of the initializers. That is in this case the array will have only 4 elements.

However if you would write the following way

char x[5] = {'W', 'o', 'r', 'd'};

then the fifth element will contain the terminating zero because it has no corresponding initializer and will be zero-initialized.

Also take into account that the following record is valid in C but invalid in C++

char x[4] = "Word";

Upvotes: 5

Carl
Carl

Reputation: 44438

No, it doesn't. If you want it inserted automatically, use a string constant:

const char *x = "Word";
//OR 
std::string const s = "Word";
const char *x = s.c_str();
//OR
char x[] = { "Word" }; //using the C++11 brace initializer syntax.

The simplest way to verify this is to look at the memory address for x using a debugger and monitor it and the next few bytes following it. Stop the debugger after the initialization and examine the memory contents.

Upvotes: 1

Jens Gustedt
Jens Gustedt

Reputation: 78903

No, if you want to use the char array as a string use this form of the initializer

char x[] = { "Word" };

Upvotes: 0

Fahad Siddiqui
Fahad Siddiqui

Reputation: 1849

No, in this case you are just creating a non-null terminated array. In-case of a read based on null termination. Your program won't stop as it wouldn't find the null at last.

char x[] = "Word";

In this case 5 bytes would be allocated for x. Null at the end.

Upvotes: 1

David Rodr&#237;guez - dribeas
David Rodr&#237;guez - dribeas

Reputation: 208323

That is undefined behavior, and it printed "Word" as it might also have crashed.

Upvotes: 3

Melika Barzegaran
Melika Barzegaran

Reputation: 439

No it doesn't. You yourself have to do that if you want to use it as a string or you want to use library functions for strings for that.

Upvotes: 1

juanchopanza
juanchopanza

Reputation: 227370

No, it doesn't. Your array x has only 4 elements. You are passing std::cout a pointer to char, which is not the beginning of a null-terminated string. This is undefined behaviour.

As an example, on my platform, similar code printed Word?, with a spurious question mark.

Upvotes: 6

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