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);
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.
width and text_length are expected to be the same in this case unless there is a rare print error.The manual page of sprintf() says
int sprintf(char *str, const char *format, ...);
sprintf(), write to the character stringstr.
Here
int value = sprintf(array, "%d", input);
sprintf() converts the int input into char array.
For e.g 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
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);
scanf()converts both01and1into the same bit pattern (a bunch of zero bits and a final one bit) — andsprintf()cannot tell how many leading zeros were in the text that was converted (if, indeed, any text was converted). Note that in source code,0777and777produce very different values; the first is an octal constant, the second a decimal constant. In decimal input withscanf("%d", …);, the leading zeros are ignored; withscanf("%i", …);, the leading zeros matter (it is octal again).01converted to an int a different number then1? What is theexact valueof01when converted to an int ?sprintf seems to ignore 0. You wanthow to use scanf to extract all numbers by their position in string.