Whats wrong with this C code?

Question

My sourcecode:

#include <stdio.h>

int main()
{
    char myArray[150];
    int n = sizeof(myArray);
    for(int i = 0; i < n; i++)
    {
        myArray[i] = i + 1; 
        printf("%d\n", myArray[i]); 
    }
    return 0;
}

I'm using Ubuntu 14 and gcc to compile it, what it prints out is:

1
2
3
...
125
126
127
-128
-127
-126
-125
...

Why doesn't it just count up to 150?

Answer 1

int value of a char can range from 0 to 255 or -127 to 127, depending on implementation. Therefore once the value reaches 127 in your case, it overflows and you get negative value as output.

Answer 2

The signedness of a plain char is implementation defined.

In your case, a char is a signed char, which can hold the value of a range to -128 to +127.

As you're incrementing the value of i beyond the limit signed char can hold and trying to assign the same to myArray[i] you're facing an implementation-defined behaviour.

To quote C11, chapter §6.3.1.4,

3. Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

Answer 3

Because a char is a SIGNED BYTE. That means it's value range is -128 -> 127.

EDIT Due to all the below comment suggesting this is wrong / not the issue / signdness / what not...

Running this code:

char a, b;
unsigned char c, d;
int si, ui, t;
t = 200;

a = b = t;
c = d = t;
si = a + b;
ui = c + d;

printf("Signed:%d | Unsigned:%d", si, ui);

Prints: Signed:-112 | Unsigned:400
Try yourself

The reason is the same. a & b are signed chars (signed variables of size byte - 8bits). c & d are unsigned. Assigning 200 to the signed variables overflows and they get the value -56. In memory, a, b,c&d` all hold the same value, but when used their type "signdness" dictates how the value is used, and in this case it makes a big difference.

Note about standard

It has been noted (in the comments to this answer, as well as other answers) that the standard doesn't mandate that char is signed. That is true. However, in the case presented by OP, as well the code above, char IS signed.

Answer 4

It seems that your compiler by default considers type char like type signed char. In this case CHAR_MIN is equal to SCHAR_MIN and in turn equal to -128 while CHAR_MAX is equal to SCHAR_MAX and in turn equal to 127 (See header <limits.h>)

According to the C Standard (6.2.5 Types)

15 The three types char, signed char, and unsigned char are collectively called the character types. The implementation shall define char to have the same range, representation, and behavior as either signed char or unsigned char

For signed types one bit is used as the sign bit. So for the type signed char the maximum value corresponds to the following representation in the hexadecimal notation

0x7F

and equal to 127. The most significant bit is the signed bit and is equal to 0.

For negative values the signed bit is set to 1 and for example -128 is represented like

0x80

When in your program the value stored in char reaches its positive maximum 0x7Fand was increased it becomes equal to 0x80 that in the decimal notation is equal to -128.

You should explicitly use type unsigned char instead of the char if you want that the result of the program execution did not depend on the compiler settings.

Or in the printf statement you could explicitly cast type char to type unsigned char. For example

printf("%d\n", ( unsigned char )myArray[i]); 

Or to compare results you could write in the loop

printf("%d %d\n", myArray[i], ( unsigned char )myArray[i]);