将Int转换为char *指针

Question

I'm trying to convert an integer to a char* pointer so I can properly send the argument into another function. Is there anyway to do this without atoi?

int number=2123, number2= 1233; 
char* arg[];

arg[0] = number;
arg[1] = number2;

**Sorry for not making things clear so basically I want the arg[0] to equal the string rep of the number, so that i can send it into a function like this: converted(char* arg);

i would change the parameter data type but it has to be that pointer to send in.

Answer 1

to convert an integer to a char* pointer so I can properly send the argument into another function.

Create a string representation of the number by calling a helper function with a char array via a compound literal to convert the number. The string will be valid for the function and to the end of the block of code. Adequate sizing of the string is important - leave that for a separate exercise. Code here uses 41, enough for 128 bit int . No need for a memory management via malloc()/free() for such a small buffer.

char *my_itoa(char *dest, int i) {
  sprintf(dest, "%d", i);
  return dest;
}

#define ITOA(n) my_itoa((char [41]) { 0 }, (n) )

int main(void) {
  int number=2123, number2= 1233;
  printf("<%s> <%s>\n",  ITOA(number),  ITOA(number2));
  printf("<%s> <%s> <%s>\n",  ITOA(INT_MIN),  ITOA(0),  ITOA(INT_MAX));
}

Output

<2123> <1233>
<-2147483648> <0> <2147483647>

Answer 2

If you assume that a pointer is always at least the same size as an integer (are there any exceptions to this?) you can safely convert integers to pointers and back with these macros:

#define INT2POINTER(a) ((char*)(intptr_t)(a))
#define POINTER2INT(a) ((int)(intptr_t)(a))

Or if that other function isn't your function, but a function that wants the integers as string you could use asprintf() like this:

asprintf(&arg[0],"%d",number);
asprintf(&arg[1],"%d",number2);

But asprintf is not posix standard so it might not be available on all systems. You could use sprintf() (or snprintf() so be on the safe side) instead, but then you need to calculate the string length first.

Answer 3

Here is the function you"re looking for:

char    *int_to_char(int nb)
{
  int   div;
  char  *str;
  int   i;
  int   size;
  int   sign;

  i = -1;
  sign = 1;
  size = get_int_lenght(nb);
  if (!(str = malloc(sizeof(char) * (size + 1))))
    return (NULL);
  if (nb < 0)
    {
      sign *= -1;
      nb *= -1;
    }
  if (nb == 0)
    {
      str[++i] = '0';
      str[i + 1] = 0;
      return (str);
    }
  while (nb > 0)
    {
      str[++i] = (nb % 10) + '0';
      nb /= 10;
    }
  if (sign < 0)
    str[++i] = '-';
  str[i + 1] = 0;
  return (my_revstr(str));
}

With revstr function:

char    *my_revstr(char *str)
{
  int   i;
  int   j;
  char  tmp;

  i = -1;
  j = strlen(str);
  while (++i < j)
    {
      tmp = str[i];
      str[i] = str[--j];
      str[j] = tmp;
    }
  return (str);
}

And get_int_lenght function:

int     get_int_lenght(int nb)
{
  int   size;

  size = 0;
  if (nb <= 0)
    {
      size++;
      nb *= -1;
    }
  while (nb > 0)
    {
      size++;
      nb /= 10;
    }
  return (size);
}