# Standard Inputs of C

Review of common inputs: scanf, getchar, gets (fgets).

## Basic Formats

1. What is the result of the following code?
#include <stdio.h>

int main(void) {
int a, b;
scanf("%d%d", a, b);
printf("%d %d", a, b);
return 0;
}


Input:

12 34


Click to expand Answer

Undefined Behavior.

scanf requires the memory address of the variable.

2. What is the result of the following code?
#include <stdio.h>

int main(void) {
int a, b;
scanf("%1d%2d", &a, &b);
printf("%d %d", a, b);
return 0;
}


Input:

1234 5678

Click to expand Answer

1 23


The %2d part means to read in 2 digits.

3. What is the result of the following code?

#include <stdio.h>

int main(void) {
int a, b;
scanf("(%d,%d)", &a, &b);
printf("%d %d", a, b);
return 0;
}


Input:

(12,34)


Click to expand Answer

12 34


What if the input becomes:

(12, 34)


Click to expand Answer

12 34


If we specify what certain character to match in scanf, the input must input that specific character. For example, using ( in scanf indicates the next character must be (, the call will fail if the next character is space or \n, … However, using %d, %f, … automatically ignores every whitespace characters (space, \n, …) before the number digits. %c is an exception that does read in whitespace characters.

4. What is the result of the following code?
#include <stdio.h>

int main(void) {
float f;
double d;
scanf("%f%f", &f, &d);
printf("%f %f", f, d);
return 0;
}

Click to expand Answer

Undefined Behavior.

The input format should be %lf for double variables, using %f will leave half of the variable’s memory undefined.

## Usages of scanf("%c", ...) and getchar

1. What is the result of the following code?
#include <stdio.h>

int main(void) {
int i;
char a[6];
for (i = 0; i < 6; i++)
scanf("%c", &a[i]);
for (i = 0; i < 6; i++)
printf("%d ", (int)a[i]);
return 0;
}


Input:

A B
C



Hint: 'A' is 65, ' ' is 32, '\n' is 10.

Click to expand Answer

65 32 66 10 67 10


Using %c as the input format automatically reads in every character, including whitespace characters. To visualize a, it is actually {'A', ' ', 'B', ' ', '\n', 'C'}. For each variable, regardless of its type, they are stored in binary format in the memory. Take char as an example, the characters are stored by their ASCII code. It should be straightforward to see that: 'A' + ' ' is 'a'. (65 + 32 = 97)

The int cast before printing out the character is used to extend the 1-byte variable to 4-bytes to avoid undefined behavior.

2. What is the result of the following code?
#include <stdio.h>

int main(void) {
int i;
char a[6];
for (i = 0; i < 6; i++)
scanf(" %c", &a[i]);
for (i = 0; i < 6; i++)
printf("%d ", (int)a[i]);
return 0;
}


Input:

AB C
DEF G



Hint: 'A' is 65, ' ' is 32, '\n' is 10.

Click to expand Answer

65 66 67 68 69 70


The space character before %c ignores all whitespace characters.

3. What is the result of the following code?
#include <stdio.h>

int main(void) {
int i;
char a[6];
for (i = 0; i < 6; i++)
a[i] = getchar();
for (i = 0; i < 6; i++)
printf("%d ", (int)a[i]);
return 0;
}


Input:

AB C
DEF G



Hint: 'A' is 65, ' ' is 32, '\n' is 10.

Click to expand Answer

65 66 32 67 10 68


c = getchar(); is same as scanf("%c", &c);.

## Usages of scanf("%s", ...)

1. What is the result of the following code?
#include <stdio.h>

int main(void) {
int i;
char a[5] = {-1, -1, -1, -1, -1};
scanf("%s", a);
for (i = 0; i < 5; i++)
printf("%d ", (int)a[i]);
return 0;
}


Input:

 ABC DE
FG



Hint: 'A' is 65, ' ' is 32, '\n' is 10.

Click to expand Answer

65 66 67 0 -1


The value of an array variable is its memory address. So using scanf("%s", a) should be enough, of course, using scanf("%s", &a) is also OK. Using %s as the input format automatically ignores leading whitespace characters (like %d). It’ll continuously read input characters until a whitespace character, and then add a '\0' (ASCII code is 0) character indicating the termination of the string. The values after '\0' aren’t affected.

2. What is the result of the following code?
#include <stdio.h>

int main(void) {
int i;
char a[3];
scanf("%s", a);
for (i = 0; i < 3; i++)
printf("%d ", (int)a[i]);
return 0;
}


Input:

ABC



Hint: 'A' is 65, ' ' is 32, '\n' is 10.

Click to expand Answer

Undefined Behavior. (May result in a segmentation fault)

After reading ABC, scanf adds an additional '\0'. However, a[3] shouldn’t be accessed in this case. But if you tried to compile and run this code, the program may not crash most of the time, but the behavior is not defined in the C standard. This is the notorious Null-terminated string in C, which can easily lead to many bugs.

## gets vs. fgets

1. What is the result of the following code?
#include <stdio.h>

int main(void) {
int i;
char s[6] = {-1, -1, -1, -1, -1, -1};
fgets(s, 6, stdin);
for (i = 0; i < 6; i++)
printf("%d ", s[i]);
return 0;
}


Input:

ABCDEF



Hint: 'A' is 65, ' ' is 32, '\n' is 10.

Click to expand Answer

65 66 67 68 69 0


What if the input becomes:

ABC


Click to expand Answer

65 66 67 10 0 -1


fgets read in the ending newline character, and so are other whitespace characters, including the leading whitespaces!

2. What is the result of the following code?
#include <stdio.h>

int main(void) {
int i;
char s[6] = {-1, -1, -1, -1, -1, -1};
gets(s);
for (i = 0; i < 6; i++)
printf("%d ", s[i]);
return 0;
}


Input:

ABC



Hint: 'A' is 65, ' ' is 32, '\n' is 10.

Click to expand Answer

65 66 67 0 -1 -1


What if the input becomes:

ABCDEF


Click to expand Answer

Undefined Behavior. (May result in segmentation fault)

gets is like fgets but does not read in the newline character. scanf("%[^\n]", s) works the same as gets but the syntax isn’t really easy to remember.

## scanf’s Return Value

1. What is the result of the following code?
#include <stdio.h>

int main(void) {
int a, b, ret;
a = b = -1;
ret = scanf("%d%d", &a, &b);
printf("%d %d %d", ret, a, b);
return 0;
}


Input:

100 abc 1000


Click to expand Answer

1 100 -1


The return value of scanf indicates how many variables does it read successfully. Take the above code and input as an example, it reads in 100 and stores it into a, then it sees abc, which is not a number, so the matching fails. The value of b isn’t modified due to the failed matching. In C, the return value of many library functions indicates whether that function ran without error.

2. What is the result of the following code?
#include <stdio.h>

int main(void) {
int a, b, ret;
a = b = -1;
ret = scanf("(%d,%d)", &a, &b);
printf("%d %d %d", ret, a, b);
return 0;
}


Input:

( 1 , 2 )

Click to expand Answer

1 1 -1


For ,, whitespace characters are invalid, so the match fails.

## End of File EOF

EOF is defined to be -1 in most cases. When the input ends, using scanf, getchar result in EOF.

For gets and fgets, they return NULL, which is 0 in most cases.

To simulate EOF, press Ctrl+Z and Enter on Windows; press Ctrl+D and Enter on macOS or Linux.

For some common usages:

while (scanf("%d", &x) != EOF) {
// Do something.
}
while (scanf("%d%d", &a, &b) == 2) {
// Do something.
}
while ((c = getchar()) != EOF) {
// Do something.
}
// Change 'MAX_STRLEN' to a constant (max string length including null-terminator)
while (fgets(s, MAX_STRLEN, stdin) != NULL) {
// Do something.
}


## Case Study

There are often multiple ways to read a certain input.

If we want to read in each of the digits of a 5-digit number:

12345



And store them into 5 variables a, b, c, d, e respectively.

1. Math
int n;
scanf("%d", &n);
a = n / 10000 % 10;
b = n / 1000 % 10;
c = n / 100 % 10;
d = n / 10 % 10;
e = n / 1 % 10;

2. Scan 1-digit at a time
scanf("%1d%1d%1d%1d%1d", &a, &b, &c, &d, &e);

3. Scan by char
a = getchar() - '0';
b = getchar() - '0';
c = getchar() - '0';
d = getchar() - '0';
e = getchar() - '0';

4. Scan by string
char str[6];
scanf("%s", &str);
a = str[0] - '0';
b = str[1] - '0';
c = str[2] - '0';
d = str[3] - '0';
e = str[4] - '0';

5. And more…

For the next assignment, we’ll review the standard outputs of C.

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