Unable to initialize c array

Question

I was writing some low level c code on fedora 14, but going nuts over this piece of code. The first array is not initialized to '0', the second is. Gone through gdb several times, but it's like magic. What is happening?

const int maxsize=100000;
char num[7];
char tnum[6];
int pos=0;
while(pos<(maxsize+1)) {
  int c=0;
  int j=0;

  int myindex;
  int tindex;
  for(myindex=0;myindex<7;myindex++) num[myindex]='0';
  for(tindex=0;tindex<6;tindex++) tnum[tindex]='0';
//....
}

I printed the array values inside gdb, both as p num , p tnum and as p num[0] and p tnum[0]. I also tried to initialize as plain 0, same thing also happens.

Here is the debugger output

Temporary breakpoint 1, main () at inversionscount.c:3
3   int main() {
Missing separate debuginfos, use: debuginfo-install glibc-2.13-1.i686
(gdb) s
5   const int maxsize=100000;
(gdb) s
6   int startarray[maxsize];
(gdb) s
14  int pos=0;
(gdb) s
15  while(pos<(maxsize+1)) {
(gdb) s
19      int c=0;
(gdb) s
20      int j=0;
(gdb) s
24      for(myindex=0;myindex<7;myindex++) num[myindex]='0';
(gdb) s
25      for(tindex=0;tindex<6;tindex++) tnum[tindex]='0';
(gdb) s
27      while( c=getchar()!="\n") {
(gdb) p num
$1 = "\370\377\277\270\367\377"
(gdb) p tnum
$2 = "000000"
(gdb)

What makes you think they aren't initialized correctly? It looks correct to me. — Paul Tomblin
– Paul Tomblin, Commented Jun 18, 2012 at 18:18
@skjaidev bzero is not a Standard C function and is deprecated in POSIX — ouah
– ouah, Commented Jun 18, 2012 at 18:22

unwind · Accepted Answer · 2012-06-18 18:16:03Z

1

What is maxsize? Make sure the code follows the execution path you think it does, by single-stepping through with a debugger.

Also, you shouldn't repeat magical constants, the for loops are better written as:

size_t i;

for(i = 0; i < sizeof num; ++i)
  num[i] = '0';
for(i = 0; i < sizeof tnum; ++i)
  tnum[i] = '0';

or, since these are char arrays, just use memset():

#include <string.h>

memset(num, '0', sizeof num);
memset(tnum, '0', sizeof tnum);

answered Jun 18, 2012 at 18:16

unwind

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3 Comments

Manlio Over a year ago

using sizeof as the exit condition of the for loop will cause the computation of the size each time. Guess it doesn't matter too much for so small arrays...

user1464460 Over a year ago

maxsize is const int with value 100000

Pedro Over a year ago

@Saphrosit: No. In all but the most convoluted cases, sizeof is evaluated at compile-time.

ouah · Accepted Answer · 2012-06-18 18:16:26Z

0

How do you check you array are initialized with the '0' character literal? If you print them as strings, remember string are null terminated and your arrays are not.

Add this:

 num[sizeof num - 1] = '\0';
 tnum[sizeof tnum - 1] = '\0';

before printing them:

printf("%s\n", num);
printf("%s\n", tnum);

Also notice '\0' is the int value 0 while '0' is the printable character 0.

answered Jun 18, 2012 at 18:16

ouah

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3 Comments

user1464460 Over a year ago

Yes your solution works. Print f prints both as (gdb) s 29 printf("%s\n", num); (gdb) s 000000 30 printf("%s\n", tnum); (gdb) s 00000 Was that a string terminating issue?Thank you.

ouah Over a year ago

@user1464460 By definition a string in C is a null terminated array of characters. The null character '\0' marks the end of the string.

user1464460 Over a year ago

Initally the problem happened when i gave input 0, mistakenly taking it for int, while was char. C needs much attention at these things :-)

blearn · Accepted Answer · 2012-06-18 18:18:18Z

0

You should use memset for both of your arrays:

#include <string.h>
char num[7];
memset(num, '\0', sizeof(num));

edited Jun 18, 2012 at 18:18

answered Jun 18, 2012 at 18:17

blearn

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2 Comments

Alex Lockwood Over a year ago

This probably isn't the solution he is looking for.

blearn Over a year ago

Why? This is the standard idiom to initialize arrays in c.

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