如何在 C 中复制一个字符数组？

Question

In C, I have two char arrays:

char array1[18] = "abcdefg";
char array2[18];

How to copy the value of array1 to array2 ? Can I just do this: array2 = array1 ?

Answer 1

You can't directly do array2 = array1 , because in this case you manipulate the addresses of the arrays ( char * ) and not of their inner values ( char ).

What you, conceptually, want is to do is iterate through all the chars of your source ( array1 ) and copy them to the destination ( array2 ). There are several ways to do this. For example you could write a simple for loop, or use memcpy .

That being said, the recommended way for strings is to use strncpy . It prevents common errors resulting in, for example, buffer overflows (which is especially dangerous if array1 is filled from user input: keyboard, network, etc). Like so:

// Will copy 18 characters from array1 to array2
strncpy(array2, array1, 18);

As @Prof. Falken mentioned in a comment, strncpy can be evil . Make sure your target buffer is big enough to contain the source buffer (including the \\0 at the end of the string).

Answer 2

如果您的数组不是字符串数组，请使用： memcpy(array2, array1, sizeof(array2));

Answer 3

If you want to guard against non-terminated strings, which can cause all sorts of problems, copy your string like this:

char array1[18] = {"abcdefg"};
char array2[18];

size_t destination_size = sizeof (array2);

strncpy(array2, array1, destination_size);
array2[destination_size - 1] = '\0';

That last line is actually important, because strncpy() does not always null terminate strings. (If the destination buffer is too small to contain the whole source string, sntrcpy() will not null terminate the destination string.)

The manpage for strncpy() even states "Warning: If there is no null byte among the first n bytes of src, the string placed in dest will not be null-terminated."

The reason strncpy() behaves this somewhat odd way, is because it was not actually originally intended as a safe way to copy strings.

Another way is to use snprintf() as a safe replacement for strcpy():

snprintf(array2, destination_size, "%s", array1);

_{(Thanks jxh for the tip.)}

Answer 4

You cannot assign arrays, the names are constants that cannot be changed.

You can copy the contents , with:

strcpy(array2, array1);

assuming the source is a valid string and that the destination is large enough, as in your example.

Answer 5

As others have noted, strings are copied with strcpy() or its variants. In certain cases, you could use snprintf() as well.

You can only assign arrays the way you want as part of a structure assignment:

typedef struct { char a[18]; } array;
array array1 = { "abcdefg" };
array array2;

array2 = array1;

If your arrays are passed to a function, it will appear that you are allowed to assign them, but this is just an accident of the semantics. In C, an array will decay to a pointer type with the value of the address of the first member of the array, and this pointer is what gets passed. So, your array parameter in your function is really just a pointer. The assignment is just a pointer assignment:

void foo (char x[10], char y[10]) {
    x = y;    /* pointer assignment! */
    puts(x);
}

The array itself remains unchanged after returning from the function.

This "decay to pointer value" semantic for arrays is the reason that the assignment doesn't work. The l-value has the array type, but the r-value is the decayed pointer type, so the assignment is between incompatible types.

char array1[18] = "abcdefg";
char array2[18];
array2 = array1; /* fails because array1 becomes a pointer type,
                    but array2 is still an array type */

As to why the "decay to pointer value" semantic was introduced, this was to achieve a source code compatibility with the predecessor of C. You can read The Development of the C Language for details.

Answer 6

it should look like this:

void cstringcpy(char *src, char * dest)
{
    while (*src) {
        *(dest++) = *(src++);
    }
    *dest = '\0';
}
.....

char src[6] = "Hello";
char dest[6];
cstringcpy(src, dest);

Answer 7

I recommend to use memcpy() for copying data. Also if we assign a buffer to another as array2 = array1 , both array have same memory and any change in the arrary1 deflects in array2 too. But we use memcpy, both buffer have different array. I recommend memcpy() because strcpy and related function do not copy NULL character.

Answer 8

array2 = array1;

is not supported in c. You have to use functions like strcpy() to do it.

Answer 9

c functions below only ... c++ you have to do char array then use a string copy then user the string tokenizor functions... c++ made it a-lot harder to do anythng

#include <iostream>
#include <fstream>
#include <cstring>
#define TRUE 1
#define FALSE 0
typedef int Bool;
using namespace std;
Bool PalTrueFalse(char str[]);
int main(void)
{
char string[1000], ch;
int i = 0;
cout<<"Enter a message: ";

while((ch = getchar()) != '\n') //grab users input string untill 
{                               //Enter is pressed
    if (!isspace(ch) && !ispunct(ch)) //Cstring functions checking for
    {                                //spaces and punctuations of all kinds
        string[i] = tolower(ch); 
        i++;
    }
}
string[i] = '\0';  //hitting null deliminator once users input
cout<<"Your string: "<<string<<endl; 
if(PalTrueFalse(string)) //the string[i] user input is passed after
                        //being cleaned into the null function.
    cout<<"is a "<<"Palindrome\n"<<endl;
else
   cout<<"Not a palindrome\n"<<endl;
return 0;
}

Bool PalTrueFalse(char str[])
{
int left = 0;
int right = strlen(str)-1;
while (left<right)
{
   if(str[left] != str[right]) //comparing most outer values of string
       return FALSE;          //to inner values.
   left++;
   right--;
}
return TRUE;
}

Answer 10

for integer types

#include <string.h>    

int array1[10] = {0,1,2,3,4,5,6,7,8,9};
int array2[10];


memcpy(array2,array1,sizeof(array1)); // memcpy("destination","source","size")

Answer 11

You cannot assign arrays to copy them. How you can copy the contents of one into another depends on multiple factors:

For char arrays, if you know the source array is null terminated and destination array is large enough for the string in the source array, including the null terminator, use strcpy() :

#include <string.h>

char array1[18] = "abcdefg";
char array2[18];

...

strcpy(array2, array1);

If you do not know if the destination array is large enough, but the source is a C string, and you want the destination to be a proper C string, use snprinf() :

#include <stdio.h>

char array1[] = "a longer string that might not fit";
char array2[18];

...

snprintf(array2, sizeof array2, "%s", array1);

If the source array is not necessarily null terminated, but you know both arrays have the same size, you can use memcpy :

#include <string.h>

char array1[28] = "a non null terminated string";
char array2[28];

...

memcpy(array2, array1, sizeof array2);

Answer 12

Well, techincally you can…

typedef struct { char xx[18]; } arr_wrap;

char array1[18] = "abcdefg";
char array2[18];

*((arr_wrap *) array2) = *((arr_wrap *) array1);

printf("%s\n", array2);     /* "abcdefg" */

but it will not look very beautiful.

…Unless you use the C preprocessor…

#define CC_MEMCPY(DESTARR, SRCARR, ARRSIZE) \
    { struct _tmparrwrap_ { char xx[ARRSIZE]; }; *((struct _tmparrwrap_ *) DESTARR) = *((struct _tmparrwrap_ *) SRCARR); }

You can then do:

char array1[18] = "abcdefg";
char array2[18];

CC_MEMCPY(array2, array1, sizeof(array1));

printf("%s\n", array2);     /* "abcdefg" */

And it will work with any data type, not just char :

int numbers1[3] = { 1, 2, 3 };
int numbers2[3];

CC_MEMCPY(numbers2, numbers1, sizeof(numbers1));

printf("%d - %d - %d\n", numbers2[0], numbers2[1], numbers2[2]);     /* "abcdefg" */

(Yes, the code above is granted to work always and it's portable)

Answer 13

None of the above was working for me.. this works perfectly name here is char *name which is passed via the function

1. get length of char *name using strlen(name)
2. storing it in a const variable is important
3. create same length size char array
4. copy name 's content to temp using strcpy(temp, name);

use however you want, if you want original content back. strcpy(name, temp); copy temp back to name and voila works perfectly

    const int size = strlen(name);
    char temp[size];
    cout << size << endl;
    strcpy(temp, name);

Answer 14

You can't copy directly by writing array2 = array1 .

If you want to copy it manually , iterate over array1 and copy item by item as follows -

int i;
for(i=0;array1[i]!='\0';i++){
 array2[i] = array1[i]; 
}
array2[i]='\0'; //put the string terminator too

If you are ok to use string library , you can do it as follows -

strncpy ( array2, array1, sizeof(array2) );