C从字符指针初始化字符数组

Question

My question should be rather simple.

I need to give a function a char array of a pre-defined length, but I have a character pointer with variable length, but not longer than the length of my array.

Here the code:

#define SIZE_MAX_PERSON_NAME 50
char person[SIZE_MAX_PERSON_NAME];
char* currentPerson = "John";

now how would I get John into the person array but also setting the rest of the array to 0 (/NUL) ?

so that I would have

BINARY DATA: "John/NUL/NUL/NUL/NUL/NUL/NUL/NUL/NUL/NUL/NUL/NUL/NUL/NUL/NUL....."

in my memory?

sorry if this is overly stupid, but I can't seem to find a solution right now.

Answer 1

First, zero-initialize the fixed-size array :

// Using memset here because I don't know if the whole copy operation can or will be used
// multiple times. We want to be sure that the array is properly zero-initialized if the next
// string to copy is shorter than the previous.
memset(person, '\0', SIZE_MAX_PERSON_NAME);

Then, copy the variable-size string into it :

strcpy(person, currentPerson);

If you are not certain that currentPerson will fit into person :

strncpy(person, currentPerson, SIZE_MAX_PERSON_NAME - 1);

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Note that strncpy also zero-initialize the remaining bytes of the array if

strlen(currentPerson) < SIZE_MAX_PERSON_NAME - 1

So you basically have these two options :

memset(person, '\0', SIZE_MAX_PERSON_NAME);
strcpy(person, currentPerson);

Or :

strncpy(person, currentPerson, SIZE_MAX_PERSON_NAME - 1);
person[SIZE_MAX_PERSON_NAME - 1] = '\0';

Answer 2

^{After this answer was posted the question was retagged from C++ to C.}

Use a std::string , like this:

// "using namespace std;" or "using std::string;", then:

string const person = currentPerson;
old_c_function( person.c_str() );

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To do things at the C level, which I recommend that you don't, first replace the unnecessary #define with a typed constant:

int const max_person_name_size = 50;

Then zero-initialize your array:

char person[max_person_name_size] = {};

(Note: no silly memset here.) (Also note: this zeroing is only a preventive measure. You wanted it. But it's not really necessary since strcpy will ensure a trailing zero-byte.)

Then just copy in the string:

assert( strlen( current_person ) < max_person_name_size );
strcpy( person, current_person );

But don't do this. Use std::string instead.

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Update : doing other things for some minutes made me realize that this answer, as all the others so far, is completely off the mark . The OP states in a comment elsewhere that

” I've got a function in the library which only takes a character array. Not a character pointer.

Thus, apparently it's all about a misconception.

The only way this can make sense is if the array is modified by the function, and then std::string::c_str() is not a solution. But a std::string can still be used, if its length is set to something sufficient for the C function. Can go like this:

person.resize( max_person_name_size );
foo( &person[0] );                // Assuming foo modifies the array.
person.resize( strlen( person.c_str() ) );

Answer 3

With literal, you may do:

char person[SIZE_MAX_PERSON_NAME] = "John";

if c-string is not a literal, you have to do the copy with strcpy

strcpy(person, currentPerson);

Answer 4

This is the one and only reason for the existence of strncpy :

Putting a string (up to the 0-terminator or buffer end) into a fixed-length array and zeroing out the rest.
This does not ensure 0-termination, thus avoid it for anything else.

7.24.2.4 The strncpy function

 #include <string.h> char *strncpy(char * restrict s1, const char * restrict s2, size_t n); 

2 The strncpy function copies not more than n characters (characters that follow a null character are not copied) from the array pointed to by s2 to the array pointed to by s1 . ₃₀₈₎ If copying takes place between objects that overlap, the behavior is undefined.
3 If the array pointed to by s2 is a string that is shorter than n characters, null characters are appended to the copy in the array pointed to by s1 , until n characters in all have been written.
4 The strncpy function returns the value of s1 .