CareerCup

very easy :)
int atoi(char *n)
{
int num;
while(*n!='\0'&&(*n>='0' && *n<='9'))
{num=num*10+*n-'0';
n++;
}
return num;

}

int main()
{
char *str = "1234";
int n;
n=myatoi(str);
printf("%d\n", n);
return 0;
}

myatoi(char *ptr)
{
int val=0;
if(ptr)
{
while(*ptr && (*ptr > '0' || *ptr <'9'))
{
val= (val * 10) + (*ptr - '0');
ptr++;
}
}
return val;
}

int atoi(const char* string)
{
    char c;
    int result = 0;
    int sign;
    if(*string == '-')
        {
        sign = -1;
        string++;
    }
    else
        sign = 1;
    while((c = *string++) != '\0')
        result = result*10+(c-'0');
    return result*sign;
}

1. You could cause overflow in result = result*10+(c-'0');
2. Check whether each character in the string could be converted to integer

<pre lang="" line="1" title="CodeMonkey57626" class="run-this">main()
{
string str="1234"
int num=0,i;
for(i=0;i<strlen(str);i++)
{

num=(num*10)+(str[i]-'0');
}
}

</pre><pre title="CodeMonkey57626" input="yes">
</pre>

///#include <stdio.h>
#include <string.h>

int main()
{
int no;
char *str="-47978757";
no=convert(str);
printf("%d",no);
return 0;
}

int convert(char *str)
{

int k=1;
int i;
int sum=0;

for(i=strlen(str)-1;i>=1;i--)
{

sum+=(str[i]-'0')*k;
k=k*10;

}
if(str[0]!='-')
sum+=(str[0]-'0')*k;
else
sum=sum*(-1);

return sum;
}\\\

#include <stdio.h>
#include <string.h>

int main()
{
    int no;
    char *str="-47978757";
    no=convert(str);
    printf("%d",no);
    return 0;
}

int convert(char *str)
{
    
    int k=1;
    int i;
    int sum=0;
    
    for(i=strlen(str)-1;i>=1;i--)
    {
        
        sum+=(str[i]-'0')*k;
        k=k*10;
      
    }
    if(str[0]!='-')
    sum+=(str[0]-'0')*k;
    else
    sum=sum*(-1);
    
    return sum;
}