Exercise 5.3 - strcat(s,t) copies the string t to the end of s¶

Question¶

Write a pointer version of the function strcat that we showed in Chapter 2: strcat(s,t) copies the string t to the end of s.

#include<stdio.h>

#define MAXLINE 1000

int mgetline(char line[], int lim);

void mystrcat(char *, char *);

int main(void) {
    int len;
    char s[MAXLINE], t[MAXLINE];

    putchar('s');
    putchar(':');
    mgetline(s, MAXLINE);

    putchar('t');
    putchar(':');
    mgetline(t, MAXLINE);

    mystrcat(s, t);

    printf("%s", s);

    return 0;
}

int mgetline(char line[], int lim) {
    int c, i;

    for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
        line[i] = c;

    if (c == '\n') {
        line[i] = c;
        ++i;
    }

    line[i] = '\0';

    return i;
}

void mystrcat(char *s, char *t) {
    while (*s != '\0')
        s++;
    s--;             /* goes back to \0 char */
    while ((*s = *t) != '\0') {
        s++;
        t++;
    }
}

Explanation¶

This is a string concatenation program using pointers. The function mystrcat is defined to take two strings as character pointers mystrcat(char *s, char *t) and this function returns the concatenated string in s itself.

The way it does is, the position in s is advanced till we meet a 0 character and then we append the characters from the string t to s, starting from the 0 character till we hit the end of the string t which is a 0 again.

void mystrcat(char *s,char *t)
{
        while(*s!='\0')
            s++;
        s--;                      /* goes back to \0 char */
        while((*s=*t)!='\0')
        {
           s++;
           t++;
        }
}

The construct while((*s=*t)!=’0’) assigns the character in t to s and then checks if the character is 0.