如何在没有字符串函数的情况下将新字符串分配给char数组

Question

I am looking to modify a char array with different strings, like a temp char array which takes various strings. Let's say a char array A[10] = "alice", how to assign A[10] = "12". Without using string functions?

TIA

Answer 1

In C, a string is just an array of type char that contains printable characters followed by a terminating null character ( '\\0' ).

With this knowledge, you can eschew the standard functions strcpy and strcat and assign a string manually:

A[0] = '1';
A[1] = '2';
A[2] = '\0';

If there were characters in the string A beyond index 2 , they don't matter since string processing functions will stop reading the string once they encounter the null terminator at A[2] .

Answer 2

it's like Govind Parmar's answer but with for loop.

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char str[11] = "hello world";
    char new[5] = "2018";   
    int i = 0;

    for (i; new[i] != '\0'; i++)
         str[i] = new[i];

    str[i] = '\0';

    printf("str => '%s' ",str);

    return 0;
}

output :

str => '2018'                                                                                                              

Answer 3

Well since string array is noting but pointer to array you can simply assign like this

int main(void) {
    char *name[] = { "Illegal month",
                            "January", "February", "March", "April", "May", "June",
                            "July", "August", "September", "October", "November", "December"
    };
    name[10] = "newstring";
    printf("%s",name[10]);
    return 0;
}