使用sprintf将整数转换为字符串似乎忽略了0

Question

When I run the following code with an input of '01', value has the value of 1, ignoring the 0. However if i input '301' with the 0 not in the first position the code works.

int input;
printf("Enter an number: \n");
scanf(" %d", &input);

char array[10];
int value = sprintf(array, "%d", input);
printf("%d", value);

Answer 1

A way to discern leading '0' digits is to record the scan offset of input.

int n1, n2;
int input;
if (scanf(" %n%d%n", &n1, &input, &n2) == 1) {
  char array[40];
  int width = n2 - n1;
  int text_length = sprintf(array, "%0*d", width, input); // "0" --> Pad with zeros
                                                       // "*" --> Min width in argument list
  printf("%d <%s>\n", text_length, array);  
}

" " will consume leading white-space. "%n" will record the scan text offset position. "%n" does not contribute to the scanf() return value.

Limitations: input like "-123" and "+123" will report 4.

Answer 2

The manual page of sprintf() says

 int sprintf(char *str, const char *format, ...); 

sprintf() , write to the character string str .

Here

int value = sprintf(array, "%d", input);

sprintf() converts the int input into char array .

For eg if user entered input as integer 123 it conver that into char array 123 . Now it looks like

-------------------------
|  1  |  2  |  3  |  \0  |
-------------------------
array

and sprintf() returns return the number of characters printed (excluding the null byte used to end output to strings). That means

int value = sprintf(array, "%d", input); /* if input = 123(integer) */
printf("%s: ,%d: \n", array,value);/* array: 123(string), value: 3 */

When I run the following code with an input of '01', value has the value of 1, ignoring the 0. ? input is declared as an integer and when user gives 01 then scanf() considers only 1 as leading 0 ignored and only 1 gets stored into array , the array looks like

--------------
|  1  |  \0  |
--------------
array

However if i input '301' with the 0 not in the first position the code works. If user entered 301 then scanf() stored 301 into input and sprintf() converts that int into char array & stores into array as 301 like

     -------------------------
    |  3  |  0  |  1  |  \0  |
    -------------------------
    array

Answer 3

When you read a string as a number using the %d format specifier, any leading zeros are basically lost.

If you want to keep the leading zeros, you need to read the input as a string. That way you get exactly what the user entered. If you additionally want to perform numerical operations, you can then use strtol to get the numerical representation.

char array[10];
scanf("%s", array);

int input = strtol(array, NULL, 10);
printf("input as number %d, input as string: %s\n", input, array);