Standard Inputs of C

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

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Update (2020.11.25): Fix typo in answer description (Question 5)

Basic Formats

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

✔️ Answer

Undefined Behavior.

scanf requires the memory address of the variable.

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

✔️ Answer

1 23

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

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
✔️ Answer
12 34
What if the input becomes:
```
(12, 34)
```
Click to expand Answer
✔️ Answer
12 34
If we specify a certain character to match in scanf, the input must include 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.
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

✔️ 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`

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
✔️ Answer
65 32 66 10 67 10
Using %c as the input format automatically reads in every character, including whitespace characters. The visualization of variable a is: {'A', ' ', 'B', '\n', 'C', '\n'}. For each variable, they are stored in its binary format in the memory, regardless of its type. 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.

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

✔️ Answer

65 66 67 68 69 70

The space character before %c ignores all whitespace characters.

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

✔️ Answer

65 66 32 67 10 68

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

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

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
✔️ 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 valid.

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 0) character indicating the termination of the string. The values after '\0' aren’t affected.
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

✔️ Answer

Undefined Behavior. (May result in a segmentation fault (SEGFAULT))

After reading ABC, scanf adds an additional '\0' at the end of the input. However, a[3] shouldn’t be accessed in this case. 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 if the programmer is not careful enough.

`gets` vs. `fgets`

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 ", (int)s[i]);
    return 0;
}

Input:

ABCDEF

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

Click to expand Answer

✔️ Answer

65 66 67 68 69 0

What if the input becomes:

AB C

Click to expand Answer

✔️ Answer

65 66 32 67 10 0

fgets reads in the ending newline character, so as other whitespace characters, including leading whitespaces!

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:
```
AB C
```
Hint: 'A' is 65, ' ' is 32, '\n' is 10.
Click to expand Answer
✔️ Answer
65 66 32 67 0 -1
What if the input becomes:
```
ABCDEF
```
Click to expand Answer

✔️ Answer

Undefined Behavior. (May result in a 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 may be difficult to memorize. %[...] tells scanf to read any number of the characters specified in the brackets. ^ inside the brackets indicates negation. So %[^\n] tells scanf to read all characters that aren’t the newline character \n.

`scanf`’s Return Value

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
✔️ Answer
1 100 -1
The return value of scanf indicates the number of variables that has been 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.

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

✔️ 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 of the compiler implementations. When the input ends, scanf and getchar returns EOF.

For gets and fgets, they return NULL, which is 0 in most of the compiler implementations.

After interacting with a program on the command line / terminal, the user can tell the program that there will be no further inputs by signaling EOF. EOF is sent by pressing Ctrl+Z and Enter on Windows or pressing 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.
}

A Toy Example

For reading a particular input format, there exist many different code snippets that can achieve the same result. However, one might be simpler than another.

For example, if we want to read a 5-digit number, with each of its digit stored in a different variable:

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

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;

Scan 1-digit at a time

scanf("%1d%1d%1d%1d%1d", &a, &b, &c, &d, &e);

Scan by char

a = getchar() - '0';
b = getchar() - '0';
c = getchar() - '0';
d = getchar() - '0';
e = getchar() - '0';

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';

And more…

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

Epilogue

A piece of code that works on your computer doesn’t necessarily mean that it will also run successfully on another person’s computer. You should be able to identify and fix codes with undefined behaviors after finishing this practice.

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Basic Formats

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

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

gets vs. fgets

scanf’s Return Value

End of File EOF

A Toy Example

Epilogue

Comments

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

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

`gets` vs. `fgets`

`scanf`’s Return Value

End of File `EOF`