我正在尝试使用指针在c中复制字符串

Question

char ch0[10] = "hello";
char ch1[10];

void main(){
    clrscr();
    char *pt0 = ch0;
    char *pt1 = ch1;
    puts(pt0);

    while(*pt0 != '\0')
    {
        *pt1++ = *pt0++;
    }
    *pt1 = '\0';

    printf("value of ch1 =");
    for(int i = 0; i < sizeof(ch1); i++){
        printf("%c",ch1[i]); // prints value correctly
    }

    putchar('\n');
    printf("pointer pt1 value = %c",*pt1); // gives garbage value
    getch();
}

Pointer pt1 value is not accessible however the ch1 is pointing to the correct value.

How to access pt1?

I am not good in pointers can any only explain me the working scenario

output:
hello
value of ch1 =hello
pointer pt1 value= \\garbage value

Answer 1

Looks to me like you just need to reset pt1 . After your copy loop, it's pointing to the end of the string in the ch1 array.

So, after the line:

*pt1='\0';

pt1 is pointing to the end of the string in ch1 . So, to print it out, you need to reset it back to ch1 .

Answer 2

There is a typo in your code:

char *pt0 =ch;

should perhaps be:

char *pt0 =ch0;

Your compiler should notice that but maybe it would be better to correct it here also for consistency. Apart from this the bruceg's answer is right.

Answer 3

In your code, you are incrementing the pointer ptr1 and then atlast your assigning value "NULL" . So no longer the ptr1 holds the ch1 string.

For better understanding, you can print the addresses of both ch1 and ptr1 . You will understand much better