This question already has answers here:
Is char signed or unsigned by default?
(6 answers)
Closed 4 years ago.
#include <stdio.h>
int main()
{
int i = 23;
char c = -23;
if (i < c)
printf("Yes\n");
else
printf("No\n");
}
This segment of code executes the else block, but it executes the if block if we replace integer value with unsigned int i = 23.
Printing these values with %x results in
c=ffffffe9
i=17
My question here is how unsigned ints and char variables gets stored in memory?
When you compare two different types, your compiler will perform a cast to the most suitable one. But in your case, there is no good solution: either you loose the sign, or the precision.
What is happening is that your compiler is promoting your char to unsigned byte.
You should check if your char is negative prior comparing it to an unsigned to prevent this from happening.
Answering what you are actually asking, a negative character is stored in memory using its two's complement.
This should help you understand it a bit better: Signed to unsigned conversion in C - is it always safe?
"The int data type is signed and has a minimum range of at least -32767 through 32767 inclusive. The actual values are given in limits.h as INT_MIN and INT_MAX respectively.
An unsigned int has a minimal range of 0 through 65535 inclusive with the actual maximum value being UINT_MAX from that same header file."
Enjoy! :)
Related
This question already has answers here:
Int to char conversion rule in C when int is outside the range of char
(2 answers)
Closed last year.
#include <stdio.h>
#include <stdlib.h>
int main()
{
signed char chr=128;
printf("%d\n",chr);
return 0;
}
Do you know about integer limits? A char value takes up 1 byte. A byte is usually 8 bits. To calculate the limit through the number of bits, the calculation is 2^n-1 meaning an integer with 8 bits has a range from 0 to 255 when unsigned. Since your variable is signed, it allocates a bit to the sign, meaning it has a range from -128 to 127. Since you assigned it as 128, it overflowed, rolling back over to -128. If your program doesn't use negative numbers, you should use signed char, otherwise you might want to use a short which is 2 bytes.
This question already has answers here:
Question about C behaviour for unsigned integer underflow
(3 answers)
Closed 3 years ago.
I observed that, when a unsigned char variable stores the value 0 (000000002) and it gets decremented by 1 (000000012), the variable value turns into 255 (111111112), which is the highest value that a unsigned char variable can hold.
My question is: why 000000002 - 000000012 turns into 111111112? (I want to see the arithmetic behind it)
The C code in which i observed it was this one:
#include <stdio.h>
main(){
unsigned char c = 0;
unsigned char d = c - 1;
printf("%d\n%d", c, d);
}
When it runs, the following output is shown:
0
255
See here:
Unsigned integer arithmetic is always performed modulo 2n where n is
the number of bits in that particular integer. E.g. for unsigned int,
adding one to UINT_MAX gives 0, and subtracting one from 0 gives
UINT_MAX.
So in your example, since unsigned char is usually 8 bit, you get 28-1 = 255.
This question already has answers here:
assigning 128 to char variable in c
(3 answers)
Closed 4 years ago.
#include <stdio.h>
int main(int argc, char const *argv[])
{
int x = 128;
char y = x;
int z = y;
printf("%d\n", z);
return 0;
}
i don't understand why this program prints -128.
i have tried to convert 128 in binary base but i'm still confused on how the C compiler convert int to char and char to int.
Note : in my machine sizeof(char) = 1 and sizeof(int) = 4
Assuming a char is signed and 8 bits, its range is -128 to 127 which means the value 128 is out of range for a char. So the value is converted in a implementation-defined manner. In the case of gcc, the result is reduced modulo 28 to become in range.
What this basically means is that the low-order byte of the int value 128 is assigned to the char variable. 128 in hex as 32-bit is 0x00000080, so 0x80 is assigned to y. Assuming 2's compliment representation, this represents the value -128. When this value is then assigned to z, which is an int, that value can be represented in an int, so that's what gets assigned to it, and its representation is 0xffffff80.
C standard does not specify whether char is signed or unsigned. In fact, assigning a char type a value outside of the basic execution character set is implementation defined. You can probably use the macros in <limits.h> to verify.
I suspect on your system char is signed, which makes the max value 127. Signed interger overflow is undefined. So no guarantess on the output. In this case, it looks like it wraps around.
This question already has answers here:
int to unsigned int conversion
(7 answers)
Closed 6 years ago.
I am wondering why this code will output -5 , since they are of different types, and -5 is a very large in unsigned considering 2's complement
#include <stdio.h>
int main()
{
unsigned int x = -5; //
printf("%d", x);
}
I am very confused, how a signed int be converted into unsigned? thanks!
The initial conversion occurs when -5 is assigned to an unsigned int. At that point you have, as suggested, a very large number (the actual value depending on the number of bits in int on your machine).
The second "conversion" is really an interpretation. printf is asked to take a set of bits and print them out as though they represent an unsigned integer value. This is why the output is -5
This question already has answers here:
Unsigned and Signed int and printf
(2 answers)
Closed 6 years ago.
I know that the range of unsigned int is 0<= I <= 2^32-1
However, when I type like this in C(visual 2015)
void main(){
unsigned int k = -10;
printf("%d",k);
}
then why computer print -10 on screen?? I think there should be error.
int stores signed numbers by default, which means they can go from -2,147,483,648 to 2,147,483,647 in range. An unsigned int means you won't be using negative numbers, so the range is much larger because you have freed up the left most bit in your number which is normally used to indicate that it is signed (negative) or not. So an unsigned int can go from 0 to 4,294,967,295. This applies to types like char as well, they normally go from -128 to 127, when unsigned, a char holds one byte exactly, or 0 to 255.
Visual Studio (and most compilers) should give you a warning for trying to store a signed value into an unsigned type.
When you used printf("%d",k) The %d is telling printf() to print out a signed int. So that is what it did. If you want printf() to print out an unsigned int than you needed to use printf("%u").
in hexadecimal form, -10 is 0xFFFFFFF6
unsigned int k = -10; means
unsigned int k = 0xFFFFFFF6;
and when printing this value if you say
printf("%d",k); the compiler will evaluate the value (0xFFFFFFF6) as integer because of the %d specifier. (check http://www.cplusplus.com/reference/cstdio/printf/ about the issue)
if you say printf("%u",k); the compiler will evaluate the value (0xFFFFFFF6) as unsigned integer and will print a value between 0-2^32-1