Different results using %c and loop vs. %s in printf with null terminated string

Question

I have a variable 'jmp_code' that is declared as a char *. When I run the following commands

printf("char by char, the code is '%c%c%c%c'\n", *jmp_code, *(jmp_code+1), *(jmp_code+2),*(jmp_code+3));
printf("printing the string, the code is '%s'\n", jmp_code);

I get the following results

char by char, the code is '0,0,0, ,'
printing the string, the code is 'ö\├w≡F┴w'

I am using codeblocks. Here is the sample code I am playing with.

#include <stdio.h>
#include <string.h>

char * some_func(char * code);

char * some_func(char * code) {

    char char_array[4];

    strcpy(char_array, "000");

    code = char_array;

    return code;

}
int main ( void ) {

    char * jmp_code = NULL;

    jmp_code = some_func(jmp_code);

    printf("char by char, the code is '%c,%c,%c,%c,'\n", *jmp_code, *(jmp_code+1), *(jmp_code+2),*(jmp_code+3));
    printf("printing the string, the code is '%s'\n", jmp_code);

    return 0;

}

I am quite confused by this. Any help would be appreciated.

Thanks

I think there is something to do with the content of jmp_code. 'α' isn't a ASCII char. — wong2, Jan 14 '11 at 14:38
Amazing. char_array is on the stack of some_func() function, you feed it and than return just the pointer. As far as I can see, printf could print quite everything. Try printing the %s before %c%c%c, you probably have strange behaviours just printing the second time... — Mr.Gate, Jan 14 '11 at 15:29

score 4 · Answer 1 · edited May 23 '17 at 10:26

Some quick observations:

char * some_func(char * code) {
    char char_array[4];
    strcpy(char_array, "000");
    code = char_array;
    return code;
}

You can't assign strings using = in C. That messes things up - you're assigning code the pointer of your locally allocated char_array to code, but you're not copying the contents of the memory. Also note that since char_array is allocated on the stack (usually), you'll find it disappears when you return from that function. You could work around that with the static keyword, but I don't think that's the nicest of solutions here. You should use something along the lines of (big warning this example is not massively secure, you do need to check string lengths, but for the sake of brevity):

void some_func(char * code) {
    strcpy(code, "000");
    return;
}

(Refer to this (and this) for secure string handling advice).

And call it via some_func(jmp_code) in main. If you're not sure what this does, read up on pointers.

Second problem.

char * jmp_code = NULL;

Currently, you've declared space enough for a pointer to a char type. If you want to use my suggestion above, you'll need either to use malloc() and free() or else declare char jmp_code[4] instead, such that the space is allocated.

What do I think's happening? Well, on my system, I'm getting:

and the code is '0,0,0,,' and the code is ''

But I think it's chance that jmp_code points to the zeros on the stack provided by your some_func function. I think on your system that data has been overwritten.

Instead you're reading information that your terminal interprets as said character. Have a read of character encoding. I particularly recommend starting with The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!)

Bill Lynch · Answer 2 · 2011-01-17T00:59:03.730

2

It might be interesting to see:

const char *pos = jmp_code;
while (*pos)
    printf("%d ", *pos++);

edited Jan 17 '11 at 00:59

answered Jan 14 '11 at 15:19

Bill Lynch

80,138
16
128
173

That resulted in an infinite loop. I did add the following printf("number by number, the code is '%d,%d,%d,%d,'\n", *pos, *(pos+1), *(pos+2),*(pos+3)); which resulted in number by number, the code is '-108,92,-61,119,' – Justin Jan 14 '11 at 15:28
Sorry about the infinite loop. I had the while condition wrong. – Bill Lynch Jan 17 '11 at 00:59

score 2 · Answer 3 · answered Jan 14 '11 at 15:27

2

You're returning a reference to a temporary array. char_array goes away when some_func() retuns, but you keep using the address of it. You need to use malloc() to allocate an array and then free() it after you use it.

answered Jan 14 '11 at 15:27

Matt K

13,370
2
32
51

score 2 · Answer 4 · answered Jan 14 '11 at 15:31

2

You're printing from an invalid pointer. char_array is on the stack of some_func() function.
The function returns the pointer of something that is on the stack and will be no more after the function returns!
The first printf finds the stack still unchanged, the second, maybe, found it filled with... garbage!

answered Jan 14 '11 at 15:31

Mr.Gate

419
4
12

Thanks for the assist. The thing that confused me the most is that the garbage that was printed was always the same string 'ö\├w≡F┴w' and not something different each time. – Justin Jan 14 '11 at 20:13

score 1 · Answer 5 · answered Jan 14 '11 at 14:43

1

I think char type can not use non-ascii char codes. Meaning your string contains UTF-8 or like symbols which code could be in (0, over9000) range, while char codes could be in (0, 255) range.

answered Jan 14 '11 at 14:43

shybovycha

11,556
6
52
82

Different results using %c and loop vs. %s in printf with null terminated string

5 Answers5