Re: [avr-gcc-list] Bizarre string problem, part 3

[Eric Fu]

Andre Wrote:

----- Original Message ----- 
From: "André - BOL" <address@hidden>
To: <address@hidden>
Sent: Thursday, September 09, 2004 10:30 AM
Subject: [avr-gcc-list] Bizarre string problem, part 3


> Hi!
>
> First, I'd like to thank everybody for the tips, but they haven't
> worked. I formulated a new very simple example to demonstrate the
> problem I'm facing.
>
> Suppose I "initialize" a variable in some different and equivalent ways:
>     char a[10]  = {'T', 'T', 'T', 'T', 'T' };
>     char b[]      = "TTTTTTT";
>     char c[10];      c[0]='T';    c[1]='T';     c[2]='T';      c[3]='T';
> //i think this can't be called an initialization!!!
>     char *d      = "TTTTTTT"
>
>     now, I perform the test
>
>     if ( x[2] == 'T' ) printf("hi");    // x = a, b, c or d
>
>     All examples are LEGAL STANDARD PURE C. I chose size 10 and the 3rd
> ( [2] ! ) element to avoid avoid any doubts that I had enougth space
> reserved, and that I took the correct character. All tests should return
> true and print "hi". My example ttest.c is to be run on Pc and shows
> exactly that with printf's
>
>     PROBLEM: when using the very same code on the ATMEGA, changing the
> printf's for "turn on led 1", "turn on led 2", etc., only the third char
> sequence (C) causes correct evaluation of the if expression (i can see
> that, because I have the leds correctly connected and only the 3rd one
> lights)
>
>     I tryed this under WINDOWS AND UNDER LINUX, using avr-gcc.
>
>     Compilation and downloading commands:
>
> avr-gcc -g -mmcu=atmega32 -Wall -o t.o t.c
> avr-objcopy -j .text -O ihex t.o t.hex
> uisp -dprog=dapa --erase
> uisp -dprog=dapa --upload if=t.hex -dno-poll -v=3 --hash=32
>
>

I think the problem is that there is a diffference between " T " and ' T ',
where " T " = ' T ' + ' \0 '.
So
    char a[10]  = {'T', 'T', 'T', 'T', 'T' };    is equal to:
    char a[10]  = {' T ', ' \0 ', ' T ', ' \0 ',' T ', ' \0 ',' T ', ' \0
',' T ', ' \0 ',};    Then a[2] = ' T '.

Likewise, c[2]='T';  is equal to c[2]= ' T '; c[2]= ' \0 '; So c[2] end s up
to be ' \0 ' instead of expected ' T '.

Cheers,

Eric Fu