将const char *复制到c中的字符串

Question

In C [not C++]:

How can I copy a const char* into a string (char array)?

I have:

const char *d="/home/aemara/move/folder"
char tar[80] ;
int dl=strlen(d);
int x=0;
while (x<dl){tar[x]=d[x]; x++; }
printf("tar[80]: %s\n",tar);

Prints out: tar[80]: /home/aemara/move/folderøèB The problem is that this way, adds garbage at the end of the array [sometimes, not always] How can I fix it ? or there is another way to copy const char* into a string ?

Answer 1

strlen returns the length without the null terminator. You need to copy one more byte.

Answer 2

You forgot to add a '\\0' character at the end after copying.

To solve this, memset(tar,'\\0',80);

Or :

if(d1 < 80){ //bad idea, don't use magic numbers
  while(x < d1){ tar[x] = d[x]; x++;}
  tar[x] = '\0';
}
printf..

Answer 3

strlen 's return value doesn't include the NULL terminator.

Add the following line after the while loop

tar[dl] = '\0';

Or you could zero initialize tar when you declare the array.

char tar[80] = {0};

Now you don't need to NULL terminate after the loop.

Answer 4

This is what you should do:

const char *d="/home/aemara/move/folder";//Semi-colon was missing in your posted code
char tar[80];
memset(tar,0x00,80);//This always a best practice to memset any array before use
int dl=strlen(d);//This returns length of the string in excluding the '\0' in the string
int x=0;
if(dl<79)// Check for possible overflow, 79th byte reserved=1 byte for '\0'
while (x<dl){ tar[x]=d[x]; x++; }
if(x<80) d[x]=0;//If not using memset have to use this, xth byte initialized to '\0'
printf("\ntar[80]: %s\n",tar);