Write a program that converts a string to an integer without using standard C library functions

#include <assert.h>
#include <stddef.h>
#include <string.h>

int StringToInteger(const char *str) {
    int num = 0;
    int sign = 1;
    assert(str != NULL); //assert that str different from NULL
    if (*str == '+' || *str == '-') {
        sign = (*str == '-') ? -1 : 1;
        str++;
        return sign; // if sign start with character
    }
    
    while (*str != '
#include <assert.h>
#include <stddef.h>
#include <string.h>
int StringToInteger(const char *str) {
int num = 0;
int sign = 1;
assert(str != NULL); //assert that str different from NULL
if (*str == '+' || *str == '-') {
sign = (*str == '-') ? -1 : 1;
str++;
return sign; // if sign start with character
}
while (*str != '\0') { 
if (*str >= '0' && *str <= '9') {
num = num * 10 + (*str - '0');
} else {
return 0;
}
str++;
}
return 0;
}
') { 
        if (*str >= '0' && *str <= '9') {
            num = num * 10 + (*str - '0');
        } else {
            return 0;
        }
        str++;
    }
    return 0;
}

I tried to write a different version of the code without using the regular conversion functions but with no success.

Problems:

Do not always return 0 @tkausl

No reason to return after finding a sign @Barmar

Delete that line and later in code adjust num per sign.
Tip: also run spell checker.

      // return sign; // if sign start with charcter
// End of code
return num * sign; // Include this line
// return 0;

Advanced: avoid int overflow

• Handle the INT_MIN text like StringToInteger("-2147483648")

• Detect potential overflow before it happens. This prevents undefined behavior (UB).

#include <assert.h>
#include <errno.h>
#include <limits.h>
#include <stddef.h>
int StringToInteger(const char* str) {
assert(str != NULL);
// TBD code if desired to ignore leading white-spaces. 
int sign = *str;
if (*str == '+' || *str == '-') {
str++;
}
const char *first = s;
int num = 0;
while (*str >= '0' && *str <= '9') { 
int digit = *str - '0';
if (num <= INT_MIN/10 && (num < INT_MIN/10 || digit > -(INT_MIN%10))) {
errno = ERANGE;
num = INT_MIN;
} else {
// Sum as a negative value as the - range is more than + range.
num = num * 10 - digit;
}
str++;
}
if (*str) {
; // May want to set errno.
return 0; // Non-null character stopped loop.
}
if (s == first) {
; // May want to set errno.
return 0; // No digits.
}
if (sign != '-') {
if (num < -INT_MAX) {
num = INT_MAX;
errno = ERANGE;
} else {
num = -num;
}
}
return num;
}

The function has many return statements that do not make sense as for example

return sign; // if sign start with charcter

And your function returns either sign or 0. In fact it does not take into account the sign of the stored value in the string.

Also take into account that the string can contain leading or trailing white space characters.

You could implement the function the same way.as standard C function strtol. A simplified function can ignore the case when the stored number is too big to fit in an object of the type int.

Here is a demonstration program that shows how the function can be implemented.

#include <ctype.h>
#include <stdio.h>
int StringToInteger( const char *str )
{
const int Base = 10;
int sign = 0;
while (isspace( ( unsigned char )*str )) ++str;
if (*str == '-' || *str == '+')
{
if (*str == '-') sign = 1;
++str;
}
int value = 0;
for (unsigned char c = *str;  isdigit( c ); c = *++str )
{
int digit = c - '0';
value = Base * value + ( sign ? -digit : digit );
}
while (isspace( ( unsigned char )*str )) ++str;
return *str == '\0' ? value : 0;
}
int main( void )
{
const char *str = " -12345\t";
printf( "\"%s\" = %d\n", str, StringToInteger( str ) );
str = " 12345\t";
printf( "\"%s\" = %d\n", str, StringToInteger( str ) );
str = " 12345\tA";
printf( "\"%s\" = %d\n", str, StringToInteger( str ) );
str = " @12345\t";
printf( "\"%s\" = %d\n", str, StringToInteger( str ) );
str = " 123@45\t";
printf( "\"%s\" = %d\n", str, StringToInteger( str ) );
}

The program output is

" -12345        " = -12345
" 12345 " = 12345
" 12345 A" = 0
" @12345        " = 0
" 123@45        " = 0

Function converting string to integer (any base if you have enough digits/letters/symbols) to represent digits in this base.

int myisspace(int c)
{
switch(c)
{
case ' ':
case '\t':
case '\r':
case '\n':
return 1;
}
return 0;
}
char *mystrchr(const char *str, const char ch)
{
while(*str)
{
if(*str == ch) return str;
str++;
}
return NULL;
}
int long long cnv(const char *str, unsigned base)
{
const char digits[] = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
int sign = 1;
int long long result = 0;
if(str && (base >=2 && base < sizeof(digits)))
{
while(myisspace(*str)) str++;
if(*str == '-' || *str == '+') { sign = *str == '-' ? -1 : 1; str++;}
}
while(*str)
{
//you may check for overflow as well
//TODO integer overflow code if needed
char *rch = mystrchr(digits, *str);
if(!rch || (rch - digits) >= base)
{
if(!myisspace(*str))
{
result = 0;
break;
}
else
{
*str++;
continue;
}
}
result *= base;
result += rch - digits;
str++;
}
return result * sign;
}
int main(void)
{
printf("%lld\n", cnv("934", 10));
printf("%lld\n", cnv("934", 12));
printf("%lld\n", cnv("ffff", 16));
printf("%lld\n", cnv("01100110011", 2));
printf("%lld\n", cnv("      -01100110011", 2));
printf("%lld\n", cnv("      -42313442323", 5));
printf("%lld\n", cnv("      -42313442323      ", 12));
}