C ：我如何打印“1764 的平方根是 42 和 * 在 ascii”？

Question

ok so I am learning C and I try to use simple functions to understand basics and here I am stuck whith a segmentation fault I can't manage to make this code working h3lp please thanks you all !!!

#include <stdio.h>

#include <stdlib.h>

int     ft_sqrt(int nb) //square root
{
        unsigned int i;

        i = nb;

        while (nb < (i * i))
            i--;
        if (nb == (i * i))
            return (i);
        if (nb > (i * i))
            return (0);
}

void    ft_strcpy(char *d, char *s) // string copy
{
        while((*d++ == *s++))
            ;
}

int     ft_strlen(char *s) // string length
{
        int i = 0;
        while(s[i] != '\0')
            i++;
        return (i);
}

char    *ft_itoa(int n)  // integer to ascii
{
        char    *s;

        s = (char *)malloc(99);
        s += ft_strlen(s);
        *s = 0;
        while((*--s == n % 10 + '0') && (n /= 10))
            ;
        return (s);
}

int     ft_atoi(char *s)  //ascii to integer
{
        int i = 0;
        while(*s)
            i = 10 * i + *s++ - '0';
        return (i);
}

int     main()
{
        int     ft_sqrt(int nb);
        void    ft_strcpy(char *d, char *s);
        char    *ft_itoa(int n);
        int     ft_atoi(char *s);
        int     ft_strlen(char *s);

        int     a, *x;
        a = 0;
        char    c[40], d[4];
        c[40] = 0;
        d[4] = 0;

        a = ft_sqrt(1764);  //42 in a
        ft_strcpy(d, ft_itoa(a));  // a in d
        ft_strcpy(c, "The square root of 1764 is: ");
        x = ft_atoi(d);
        printf("\n\n\t%s%sand%cin ascii\n\n\n", c, d, x);



        return 0;
}

Just hack my code just wanna learn !!

Answer 1

There are many bugs in your code

1. When the nb is not a perfect square number , then it will give 0 always. Try it yourself. Use sqrt() instead (available under math.h header file)

2. Your ft_strcpy() is not properly framed, use strcpy() under string.h header file instead. Prototype : void strcpy(char *str1, char *str2) ,here the content of str2 will be copied to the str1 .

3. char *ft_itoa(int n) may not properly work because you have provided wrong value to ft_strlen() which may turn out wrong value to be added to the pointer *s ( Why ?. Think your self , i'm not gonna tell you that elementary concept ).

Do re-code your program and let me know if still can't fix the Error.

Answer 2

Hey @BLUEPIXY thanks for your help you fixed my atoi ! @psyco thank you here is my code fixed btw with the same value to ft_strlen() for *ft_itoa(int n).

#include <stdio.h>
#include <stdlib.h>

int     ft_sqrt(int nb)
{
        unsigned int i;

        i = nb;

        while (nb < (i * i))
            i--;
        if (nb = (i * i))
            return (i);
        if (nb > (i * i))
            return (0);
}    

void    ft_strcpy(char *d, char *s)
{
        while((*d++ = *s++))
            ;
} 

int     ft_strlen(char *s)
{
        int i = 0;
        while(s[i] != '\0')
            i++;
        return (i);
}

char    *ft_itoa(int n)
{
        char    *s;

        s = (char *)malloc(99);
        s += ft_strlen(s);
        *s = 0;
        while((*--s = n % 10 + '0') && (n /= 10))
            ;
        return (s);
}

int     ft_atoi(char *s)
{
        int i = 0;
        while(*s)
            i = 10 * i + *s++ - '0';
        return (i);
}

int     main()
{
        int     ft_sqrt(int nb);
        void    ft_strcpy(char *d, char *s);
        char    *ft_itoa(int n);
        int     ft_atoi(char *s);
        int     ft_strlen(char *s);

        int     a;
        a = 0;
        char    c[40], d[4];
        c[40] = 0;
        d[4] = 0;

        a = ft_sqrt(1764);
        ft_strcpy(d, ft_itoa(a));
        ft_strcpy(c, "The square root of 1764 is: ");
        printf("\n\n\t%s%s and %c in ascii\n\n\n", c, d, ft_atoi(d));

        return 0;
}

BugFixed thank you all