Why "initializer-string for array of chars is too long" compiles fine in C & not in C++?

Question

Following program compiles fine in C with warnings but fails in compilation in C++. Why? What is the reason?

#include <stdio.h>
int main(void)
{
    char a[5]="Hello";
    a[0]='y';
    puts(a);
    for(int i=0;i<5;i++)
        printf("%c",a[i]);
    return 0;
}

The warning:

Warning:[Error] initializer-string for array of chars is too long [-fpermissive] enabled by default

But if the program is compiled as C++ program then C++ compiler gives following error:

[Error] initializer-string for array of chars is too long [-fpermissive]

I am using GCC 4.8.1 compiler.

It gave you error on both compilers. Its just that on C++, the flag -fpermissive is lenient with the error. — Milind Dumbare, Feb 10 '15 at 14:18
You should code `char a[]="Hello";` to be sure to have a terminating 0-byte — Basile Starynkevitch, Feb 10 '15 at 14:27
**In fact the premise of this question is incorrect.** Both times you were using a **C++** compiler but once it defaulted to compile with `-fpermissive`. The `-fpermissive` is not a **GCC C compiler option at all**! — Antti Haapala -- Слава Україні, Sep 18 '19 at 07:38

score 24 · Accepted Answer · answered Feb 10 '15 at 14:26

Short answer: Because C and C++ are different languages with different rules.

Long answer: In both cases the reason is that the array is too small for the string literal. The literal consists of the five visible characters, with a zero terminator on the end, so the total size is 6.

In C, you're allowed to initialise an array with a string that's too long; extra characters are simply ignored:

C99 6.7.8/14: An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.

The compiler helpfully warns that the string is too large, since it almost certainly indicates an error; but it can't reject the code unless you tell it to treat warnings as errors.

In C++, the initialiser isn't allowed to be larger than the array:

C++11 8.5.2/2: There shall not be more initializers than there are array elements.

so, for that language, the compiler should give an error.

In both languages, when you want a character array to be the right size for a string literal initialiser, you can leave the size out and the compiler will do the right thing.

char a[] = "hello";  // size deduced to be 6

Actually this is incorrect answer. The C standard does not allow the use of too long strings. What happens here is a constraint violation for [attempting to provide a value for an object not contained within the entity being initialized.](http://port70.net/~nsz/c/c11/n1570.html#6.7.9p2). Notice that there is no length option for the successive characters of the string literal, unlike the "terminating null". — Antti Haapala -- Слава Україні, Sep 18 '19 at 07:30

Sourav Ghosh · Answer 2 · 2015-02-10T14:40:15.897

12

The problem is in below line

char a[5]="Hello";

There is no space left to store the terminating null.

By default, gcc does not produce any errors for this case with -fpermissive option enabled. So, it compiles fine in C.

As per the requirement from the language standards,

ForC99 standard, chapter 6.7.9, paragraph 14,

An array of character type may be initialized by a character string literal or UTF−8 string literal, optionally enclosed in braces. Successive bytes of the string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.

For C++11, chapter 8.5.2, paragraph 2

There shall not be more initializers than there are array elements.

So, the code compiles (with warnings) with gcc, produces error with g++.

edited Feb 10 '15 at 14:40

answered Feb 10 '15 at 14:16

Sourav Ghosh

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But i have read that if programmer doesn't specify '\0' at end of string compiler will automatically add it & it is not very serious thing. – Destructor Feb 10 '15 at 14:26
1

@meet: It's added to (or rather, included in) the string literal, yes. But it can't be added to the array of size 5, since that would require a sixth element. – Mike Seymour Feb 10 '15 at 14:27
1

@meet The space for null terminator is not there as we've defined the array to have `5` `char`s only. – Sourav Ghosh Feb 10 '15 at 14:28
@Mike Seymour: so it is allowed in C but not in C++ because C++ is a strongly typed language compared to C. right? – Destructor Feb 10 '15 at 14:29
1

@meet: It's not really got anything to do with strong typing; just different rules in different languages. C is allowed to truncate the initialiser to fit the array; C++ isn't. See my answer for the rules themselves. – Mike Seymour Feb 10 '15 at 14:30
@MikeSeymour Sir, do you mind if I add those references to my answer, too? – Sourav Ghosh Feb 10 '15 at 14:33
1

@SouravGhosh: If you like. You don't need my permission to quote the language standard. – Mike Seymour Feb 10 '15 at 14:36
2

and stop with this "Sir" stuff ;p – Lightness Races in Orbit Feb 10 '15 at 14:36

score 4 · Answer 3 · answered Feb 10 '15 at 14:17

4

A valid string in C and C++ should have a \0 terminator and in

char a[5]="Hello";

There is no space for null terminator and a is not a valid string.

So this might not be a error in C but you are bound to hit with issues when using in-built string functions like strlen() and family.

answered Feb 10 '15 at 14:17

Gopi

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This is wrong, it is perfectly fine to initialize a non-null fixed-size character array with the contents of a string literal in C. – Antti Haapala -- Слава Україні Sep 18 '19 at 07:32

score 3 · Answer 4 · answered Feb 10 '15 at 14:19

3

Because "Hello" is 6 chars long and g++ without -fpermissive won't initialise a 5 char array with it, but gcc will.

answered Feb 10 '15 at 14:19

Paul Evans

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Why C will allow this? – Destructor Feb 10 '15 at 14:19
@meet C always has and still does to remain backwards compatible. – Paul Evans Feb 10 '15 at 14:23

Why "initializer-string for array of chars is too long" compiles fine in C & not in C++?

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