I am compiling and running the following code snippet on a Linux box -
I am really
puzzled by the answers. Could someone please tell me what might be
wrong?
void test(){
int m = 0;
int n = 0;
int i = 0;
int j = 0;
double start = 0.0;
double end = 0.0;
double gap = 0.0;
struct timeval tp;
aes_context ctx;
unsigned char buf[16];
unsigned char key[32];
for(i = 0; i < 16; i++){ buf[i] = 'a'; }
for(j = 0; j < 32; j++){ key[j] = '9'; }
printf("%d, %d\n", strlen(buf), strlen(key));
/* Some more code omitted */
}
The print statements provide the answers 16, 48
Could someone please provide some hint as to what might be the problem? 5 1562 cp**********@ya hoo.com wrote: I am compiling and running the following code snippet on a Linux box - I am really puzzled by the answers. Could someone please tell me what might be wrong?
void test(){ int m = 0; int n = 0; int i = 0; int j = 0;
double start = 0.0; double end = 0.0; double gap = 0.0;
struct timeval tp;
aes_context ctx;
unsigned char buf[16];
unsigned char key[32];
for(i = 0; i < 16; i++){ buf[i] = 'a'; }
for(j = 0; j < 32; j++){ key[j] = '9'; }
printf("%d, %d\n", strlen(buf), strlen(key));
1. strlen() returns values of the type size_t; size_t is an unsigned
integer type and not necessarily the same as unsigned int.
With C89, use
printf("%lu\n", (unsigned long)strlen(som estring));
with C99, you have a separate length modifier for size_t:
printf("%zu\n", strlen(somestri ng));
2. Valid C strings are _always_ terminated by a character with the
value 0, often written as '\0' and referred to as string terminator.
Your arrays of char do not contain C strings. In order to do so, one
array element in the range 0 .. sizeof buf -1 (or sizeof key -1,
respectively) needs to be zero.
Otherwise, strlen() keeps probably reading through the memory until
it encounters a byte of value 0. Why probably? Because the behaviour
is undefined. It is also possible that the program dies with a
segmentation fault or that your computer crashes. /* Some more code omitted */ }
The print statements provide the answers 16, 48 Could someone please provide some hint as to what might be the problem?
In your case, it is possible that the first byte after the storage
reserved for buf is '\0', so you get sizeof buf. Moreover, it could
be that key is stored right before buf, so you have sizeof key plus
sizeof buf until '\0' is encountered. This is highly speculative.
To get you started:
#define SOMELEN (16)
.....
char mybuf[SOMELEN+1];
for (i=0; i<SOMELEN; i++)
mybuf[i] = '.';
mybuf[SOMELEN] = '\0';
printf("String length is %scorrect\n", strlen(mybuf)== SOMELEN
? "":"not ");
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address. unsigned char buf[16];
unsigned char key[32];
Try
unsigned char buf[16] = {};
unsigned char key[32] = {};
to initialize them to all \0s
so the strings will actually end (a string for use with the
standard libraries (calls such as strlen, printf, etc) ends at some \0)
as long as you keep some \0
and go from there
-Mysid
On 11 Feb 2005 14:58:53 -0800, cp**********@ya hoo.com wrote in
comp.lang.c: I am compiling and running the following code snippet on a Linux box - I am really puzzled by the answers. Could someone please tell me what might be wrong?
void test(){ int m = 0; int n = 0; int i = 0; int j = 0;
double start = 0.0; double end = 0.0; double gap = 0.0;
struct timeval tp;
aes_context ctx;
unsigned char buf[16];
unsigned char key[32];
for(i = 0; i < 16; i++){ buf[i] = 'a'; }
for(j = 0; j < 32; j++){ key[j] = '9'; }
printf("%d, %d\n", strlen(buf), strlen(key));
Obviously you don't have a prototype for strlen() in scope, usually
done by including <string.h>, or you are operating your compiler in a
broken (non conforming) mode. Because the argument to strlen() must
be a pointer to char, and you are passing pointer to unsigned char.
No automatic conversion is possible, so each call to strlen() would be
a constraint violation.
Actually calling strlen() with a pointer to unsigned char produces
undefined behavior.
Even if you change buf and key to arrays of char, instead of unsigned
char, they are not strings, just arrays. Calling strlen() with a
pointer to char that is not a pointer to a string also produces
undefined behavior.
/* Some more code omitted */ }
The print statements provide the answers 16, 48 Could someone please provide some hint as to what might be the problem?
So...
1. Turn up the warning level on your compiler.
2. Invoke it in standard conforming mode.
3. Include <string.h> before calling strlen().
4. Do not pass pointers to unsigned char to strlen().
5. Do not pass pointers to character arrays that are not strings to
strlen().
Then your problem will go away.
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++ http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html
I would suggest to make some modifications in your code as bellow
unsigned char buf[17]; //array bound 17 to include '\0' string
termination character.
unsigned char key[33]; //Same justification as above.
for(i = 0; i < 16; i++){ buf[i] = 'a'; }
for(j = 0; j < 32; j++){ key[j] = '9'; }
buf[i] = '\0'; /*This is required as string terminates
with '\0' as you know*/
key[j] = '\0';
BTW just modifiying the code to unsigned char buf[17] = {}; will not
work because = {}; does not implicitely put '\0' at all positions of
array.
Mysidia wrote: Try
unsigned char buf[16] = {};
unsigned char buf[16] = { 0 };
unsigned char key[32] = {};
unsigned char key[32] = { 0 };
to initialize them to all \0s
There is no such thing as an empty initializer.
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