In the following code there is an argument that I cant wrap my head around. I am supposed to create a pyramid comprised letters, with each line having a unique letter. Each letter/line is in a consecutive format, so line 1 could be A and line 2 would be B B, and line 3 being C C C.
The trouble is with the first For loop that is used to create the rows of the pyramid. Here is the troublesome portion: (input-'A'+1). "input" is a user assigned character variable. There is also a character variable named "alphabet" which has 'A' assigned to it, but unless I am mistaken that is not whats being used here.
From how my brain sees 'A' in this portion of the code it is a character literal. So with this thinking I created a small program using this portion of the entire statement as a printf statement. The variable "input" is user assigned. When i run it with %c as the conversion specifier it completes with no errors but there is no visible output, and it completes with a beep. When I run it with %d it runs and gives me the integer that I expect when subtracting input from A if they were both integers.
What am i missing? is 'A' being converted to an integer? I dont understand how it works, even though I see that it does work. Here is the original code:
#include <stdio.h>
int main()
{
int i, j;
char input, alphabet='A';
printf("Enter the uppercase character you want to print in the last row ");
scanf("%c", &input);
for (i=1; i<=(input- 'A'+1); ++i) {
for(j=1; j<=i; ++j)
{ printf("%c", alphabet); }
++alphabet;
printf("\n");
}
return 0;
}
Thank you for any help in clearing this up for me as I am learning from books and YouTube.
Character literals like 'A' are actually integers. char type is signed 8 bit integer type. 'A' is equal to 65. You can check the ASCII table too see which number represent which character. With i <= (input -'A'+ 1) the code test how far is the input character from 'A' basically which is the height of the pyramid. If the input were 'A' then i <= (input -'A'+ 1) would essentially i <= (65 - 65 + 1) which is i <= 1 there for only 1 iteration would happen printing a single A.
Since char is an integer, therefore ++alphabet will gives the next character in the alphabet.
The reason you have no output with printf with %c with your testing is most likely because you try to print a non printable character. Check the ASCII table and try to print some alphabet chars or numbers by using their decimal value. Like printf("%c", 65) should output A.
This little program will print the uppercase alphabet. Might help you see things more clearly.
#include <stdio.h>
int main()
{
for (int i = 65; i < 91; i++)
printf("%d printed as character is %c\n", i, i);
return 0;
}
Output
65 printed as character is A
66 printed as character is B
67 printed as character is C
68 printed as character is D
69 printed as character is E
70 printed as character is F
71 printed as character is G
72 printed as character is H
73 printed as character is I
74 printed as character is J
75 printed as character is K
76 printed as character is L
77 printed as character is M
78 printed as character is N
79 printed as character is O
80 printed as character is P
81 printed as character is Q
82 printed as character is R
83 printed as character is S
84 printed as character is T
85 printed as character is U
86 printed as character is V
87 printed as character is W
88 printed as character is X
89 printed as character is Y
90 printed as character is Z
Characters in C are stored as integers behind the scenes and every character has a corresponding value which you can see in the ASCII table. You can always print out the character as a %d and you'll get that character's int value. You can use that as an advantage while coding so you can increment, decrement, compare and do any sort of operation you would do with basic integers.
Also, post a copy of the code you are having trouble with so others and me can look for the error.
Related
So I've seen these two questions:
Input string with non printable chars
Inputting Non ASCII characters to scanf("%s")
And unfortunately these did not provide me for an answer I've needed
let's say I'm dependent within some output of a certain program and can't tell for sure which values I need to enter ahead of program's execution, I'd rather enter some input for example for this code:
#include <stdio.h>
int main()
{
char a[10] = {0};
printf("\n----------- %x\n", (unsigned int) a);
scanf("%s", a);
for(int i = 0; i < 10; i++)
printf("%d\n", (int) a[i]);
}
as an example, I'm waiting after the printf within the following program has reached, then
input -> some input containing non-printable ascii characters
program output-> 04
08
01
05
02
10
09
12
02
02 (All these are values of non-printable ascii characters)
and again I'm looking for a way to enter them directly
I'm not looking for a way to do this by implementing some
functionality within my code, my code is given as an example of a
program
When saying during runtime, I mean after the program has started running - hence the prints within my program have reached
I apologize if this was unclaer
On Linux you can type Control-v before a control character to enter it literally. So to type the character with code 0x04, type Control-vControl-d.
my question is about how EOF is interpreted in the middle of an input, here is an example:
int main() {
int a, b;
printf("enter something >\n");
scanf("%d", &a);
while((b = getchar()) != EOF) {
printf("%i\n", b);
}
return b;
}
I run the program and enter:
1hello^Z(control+z)abc
the output is:
104 (ascii number for h)
101 (for e)
108 (l)
108 (l)
111 (o)
26 (what is this?)
The digit 1 is read by scanf, the remaining stays in the buffer, getchar() gets all of them until ^Z, which is expected behavior, as the control z closes stdin.
however where does 26 come from? If the last thing getchar() reads is EOF why isn't -1 the last value? Also why doesn't this program get out of the loop when it reads ^Z, why do I need to invoke EOF one more time with control z to terminate the loop? 26 is the ascii for SUB, I don't know what to make of this.
Thank you.
When the loop ends, b=26 because you entered ctrl+z and this is interpreted as SUB while returning .http://en.wikipedia.org/wiki/Substitute_character
In the ASCII and Unicode character sets, this character(SUB) is encoded by the number 26 (1A hex). Standard keyboards transmit this code when the Ctrl and Z keys are pressed simultaneously (Ctrl+Z, by convention often described as ^Z).
what happens if I read an integer like 20,30,10000...9999 into variable a ? it only prints the first digit in the number that I've read...why is that ?
for example if I read 123, on the screen it prints 1. Isn't it supposed to convert the integer 123 into it's equivalent ASCII character representation ?
#include <stdio.h>
int main() {
char a;
scanf("%c", &a);
printf("%c", a);
return 0;
}
This is an exam question from C language.
No, it reads the character, which is represented by the machine as a small integer, into the variable.
If you enter 100 (the number 100, three keypresses and thus three characters), it will only store the first character of that, i.e. the leading 1.
If you wanted to convert a number to an actual integer, you should use %d and an int variable of course.
Printing with %c will print back a single character, by interpreting the small integer value as a character (rather than as an integer). So for an input of 100 you will see 1 printed back out, i.e. the character that represents the decimal digit one.
If you want to print out the numeric representation of the character you read in, scan with %c but print with %d, and cast the char to (int) in the printf() call.
The problem is that %c parse a char for console input. From a number like 123 it take only the first letter and dispose the rest. The way to parse a int value is using %d on the scanf function.
No, it will read only the first character into the char variable. How can a char variable store more than one character at an instant? It can't.
So if you want the ASCII value, input as an integer instead.
int a;
scanf("%d", &a); // suppose input is 65
printf("%c", a); // prints 'A'
printf("%d", a); // prints 65
Whereas
char a;
scanf("%c", &a); // suppose input is 65
printf("%c", a); // prints '6'
printf("%d", a); // prints 54 which is the ASCII value of '6'
Practice.c
#include <stdio.h>
main()
{
char a;
printf("\nEnter Anything = ");
scanf("%c",&a); <Line 1>
printf("\n%d",a);
printf("\n%c",a);
}
Output 1 : Enter Character = 5
53
5
Output 2 : Enter Character = a
97
a
This program gets executed exactly by the book.
New.c
#include <stdio.h>
main()
{
char a;
printf("\nEnter Character = ");
scanf("%d",&a); <Line 1>
printf("\n%d",a);
printf("\n%c",a);
}
Output 1 : Enter Character = 5
5
♣
Output 2 : Enter Character = a
0
This is the same program as Practice.c with minor change in it. It is not a question but I mistakenly typed %d instead of %c in the line denoted by Line 1 of the program. This mistake of mine produced 2 very different outputs. What is the exact reason behind it?
Below ASCII 32, all are non-printable characters. In second code you are reading an int and then trying to print the equivalent character which is non-printable.
For the second input a, scanf doesn't read this character and leave it in the buffer because it expects an integer not a character. The variable a is uninitialized and you are now accessing an uninitialized variable. This invokes undefined behavior.
The character data type is having size 1 byte, whereas integer is usually 4 bytes. Here you have tried to read into a character variable an integer. There is not enough storage to handle this properly. The result depends on whether your architecture supports little endian or big endian notation.
When you use %c the input is taken as character and stored its ASCII value.
When you use %d the input is taken as integer.
When you print with %c, it tries to print ASCII char at given value.
When you print with %d, it tries to print the number.
What really happened in your NEW program is
scanf("%d",&a);
Format specifier while reading is given as an integer %d ,so the compiler assumes the input value as integer even though it is stored in char variable, so the value of variable will be
a = 0x05 which is an integer 5
printf("\n%d",a);
printing variable a as integer prints 5 on output
printf("\n%c",a);
I think you know that while reading a digit as a character(using %c) it's ASCII value is stored (for eg: scanf("%c",&a)); and you enter a as 5, the value stored in the variable a is a=0x35 in hex ,or 53 as integer representation
printing variable as character-> since 0x05 is not character or digit in ASCII value it prints the special character on the output
for about ASCII table refer : http://www.asciitable.com/
I want to type in a number that is 10 digits long, then put the digits into an array. But for some reason, I get this random 2-digit numbers that seem to have nothing to do with my input (??).
char number[10]; //number containing 10 digits
scanf("%s",number); //store digits of number
printf("%d\n",number[0]); //print the 1st digit in the number
printf("%d\n",number[1]); //print the 2nd digit in the number
Here is what I got:
Input:
1234567890
Output:
49
50
Actually, 49 should be 1, and 50 should be 2.
You are getting ASCII value of characters 1 and 2. Use %c specifier to print the digits.
printf("%c\n",number[0]);
Warning! Your code may invoke undefined behaviour!
But we'll talk about it later. Let us address your actual question first.
Here is a step by step explanation of what is going on. The first thing you need to know is that every character literal in C is actually an integer for the compiler.
Try this code.
#include <stdio.h>
int main()
{
printf("%d\n", sizeof '1');
return 0;
}
The output is:
4
This shows that the character literal '1' is represented as 4 byte integer by the compiler. Now, let us see what this 4 byte integer for '1' is using the next code here.
#include <stdio.h>
int main()
{
int a = '1';
printf("a when intepreted as int : %d\n", a);
printf("a when intepreted as char: %c\n", a);
return 0;
}
Compile it and run it. You'll see this output.
a when intepreted as int : 49
a when intepreted as char: 1
What do we learn?
The character '1' is represented as the integer 49 on my system. This is so for your system too. That's because in my system as well as yours, the compiler is using ASCII codes for the integers where '1' is 49, '2' is 50, 'A' is 65, 'B' is 66, and so on. Note that the mapping of the characters to these codes could be different for another system. You should never rely on these integer codes to identify the characters.
So when I try to print this value as integer (using %d as the format specifier), well what gets printed is the integer value of '1' which is 49. However, if we print this value as a character (using %c as the format specifier), what gets printed is the character whose integer code is 49. In other words, 1 gets printed.
Now try this code.
#include <stdio.h>
int main()
{
char s[] = "ABC123";
int i;
printf("char %%d %%c\n");
printf("---- -- --\n");
for (i = 0; i < 6; i++) {
printf("s[%d] %d %c\n", i, s[i], s[i]);
}
return 0;
}
Now you should see this output.
char %d %c
---- -- --
s[0] 65 A
s[1] 66 B
s[2] 67 C
s[3] 49 1
s[4] 50 2
s[5] 51 3
Does it make sense now? You need to use the %c format specifier when you want to print the character. You should use %d only when you want to see the integer code that represents that character.
Finally, let us come back to your code. This is how you fix it.
#include <stdio.h>
int main()
{
char number[10];
scanf("%9[^\n]", number);
printf("%c\n", number[0]);
printf("%c\n", number[1]);
return 0;
}
There are two things to note.
I have used %c as the format specifier to print the character representation of the digits read.
I have altered the format specifier for scanf to accept at most 9 characters only where the characters are not newline characters. This is to make sure that a user cannot crash your program by inputting a string that is far longer than 9 characters. Why 9 instead of 10?. Because we need to leave one cell of the array empty for the null-terminator. A longer input would overwrite memory locations beyond the allocated 10 bytes for the number array. Such buffer overruns lead to code that invoke undefined behaviour which could either cause a crash or kill your cat.
printf("%c\n",number[0]); //print the 1st digit in the number
printf("%c\n",number[1]);
should do the job for you, what you see are ascii values.
your number array is an array of char, and so every element of it is a char.
when you type:
printf("%d\n",number[0]);
you printing the chars as integers, and so you get the ASCII code for each char.
change your statement to printf("%c\n",number[0]); to print chars as chars not as ints
Warning! Your code invokes undefined behaviour!
char number[10]; // Can only store 9 digits and nul character
scanf("%s",number); // Inputting 1234567890 (11 chars) will overflow the array!
Use fgets instead:
#define MAX_LEN 10
char number[MAX_LEN];
if(fgets(number, MAX_LEN, stdin)) {
// all went ok
}
Once you have fixed this, you can fix the printing problem. You are printing the character code (number), and not the actual character. Use different type specifier:
printf("%c\n",number[0]);