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manipulating double arrays sent to function
Check the code, should be self explanitory what I'm trying to do.
The function crashes on line:
thespliced[i_word_num][i_w
Obviously because I haven't allocated memory for it.
How do allocate memory for a double array?
The function crashes on line:
thespliced[i_word_num][i_w
Obviously because I haven't allocated memory for it.
How do allocate memory for a double array?
in the parse function:
char **spliced;
return_val = splice(string,thespliced);
int splice(char *string, char *thespliced[])
{
int i_string=0;
int i_word_num=0;
int i_word=0;
int i=0;
for( ; i_string < (strlen(string) - 1) ; ++i_string)
{
if(string[i_string] == ' ')
{
++i_word_num;
i_word = 0;
continue;
}
thespliced[i_word_num][i_word] = string[i_string];
++i_word;
}
return i_word;
}
Asked:
2
2
2 Solutions
By dynamic do you mean, the array expands as you keep putting more items in it, or just that you can set the dimensions at runtime?
The first option doesn't exist in C. If you want something that expands you have to do it yourself by allocating a bigger array, copying over the data from the original smaller array into the new bigger array and then delete the original array.
The second option, you can do in 2 ways. Do you know the size of one of the dimensions? If so, you can do
char spliced[20][] = malloc(20 * xDim * sizeof(char));
then you get an array with one fixed dimension and one dimension set at runtime.
If you want both dimensions to be able to be set at runtime, then you have to change a bit more code. You can no longer use the array as thespliced[3][6], because the actual address that that refers to is calculated at compile time and if you don't know one of the dimensions, the compiler can't work it out.
To do this, you have to declare your array as
char spliced[] = malloc(xDim * yDim * sizeof(char));
and then pass it to your splice method as
int splice(char *string, char thespliced[], int xDim)
and then you access the array like...
thespliced[i * xDim + j]; // where i and j are the indexes into the array, and xDim is the size of the first dimension
Hope that helps
The first option doesn't exist in C. If you want something that expands you have to do it yourself by allocating a bigger array, copying over the data from the original smaller array into the new bigger array and then delete the original array.
The second option, you can do in 2 ways. Do you know the size of one of the dimensions? If so, you can do
char spliced[20][] = malloc(20 * xDim * sizeof(char));
then you get an array with one fixed dimension and one dimension set at runtime.
If you want both dimensions to be able to be set at runtime, then you have to change a bit more code. You can no longer use the array as thespliced[3][6], because the actual address that that refers to is calculated at compile time and if you don't know one of the dimensions, the compiler can't work it out.
To do this, you have to declare your array as
char spliced[] = malloc(xDim * yDim * sizeof(char));
and then pass it to your splice method as
int splice(char *string, char thespliced[], int xDim)
and then you access the array like...
thespliced[i * xDim + j]; // where i and j are the indexes into the array, and xDim is the size of the first dimension
Hope that helps
I think I understand your goals. I advise you to break down the problem into smaller pieces and ask separate questions until you get a better understanding of pointers and arrays.
I changed the prototypes to simplify the programming. All arrays are dynamically created. You can infer the prototypes from the usage.
One thing I did which may be of interest. I created an an extra function,
int countNumTokens( char * myString );
to return the number of space delimited words in a phrase. This function allows you to determine how much space to allocate for your top-level array of pointers. Then in each element of this array, you need to allocate space for the particular token that you parsed out of your phrase.
Here is a tutorial on pointers and arrays:
http://www.ibiblio.org/pub/languages/fortran/append-c.html
and this deals more with dynamic arrays:
http://c-faq.com/~scs/cclass/int/sx9b.html
You may wish to consider this test driver:
I recommend that you use a graphical debugger such as ddd to be able to view the memory layouts of your dynamic arrays.
I changed the prototypes to simplify the programming. All arrays are dynamically created. You can infer the prototypes from the usage.
One thing I did which may be of interest. I created an an extra function,
int countNumTokens( char * myString );
to return the number of space delimited words in a phrase. This function allows you to determine how much space to allocate for your top-level array of pointers. Then in each element of this array, you need to allocate space for the particular token that you parsed out of your phrase.
Here is a tutorial on pointers and arrays:
http://www.ibiblio.org/pub/languages/fortran/append-c.html
and this deals more with dynamic arrays:
http://c-faq.com/~scs/cclass/int/sx9b.html
You may wish to consider this test driver:
int main ()
{
char phrase[] = "A rolling stone gathers no moss";
char **tokens;
int numTokens, tokenlen, k;
tokens = parse(phrase, &numTokens );
printf("Original phrase = \"%s\"\n", phrase); // notice phrase was not altered
printf("\nNumber of words is %d\n", numTokens);
printf( "\nlook at 2nd word one char at a time\n");
tokenlen = strlen( tokens[1] );
for( k=0; k < tokenlen; ++k ) {
printf( "%c ", tokens[1][k] );
}
printf("\n\n");
printf("Word List:\n");
for( k=0; k<numTokens; ++k) {
tokenlen = strlen( tokens[k] );
printf("%2d: len=%2d: %s\n", k, tokenlen, tokens[k] );
}
freeMemory( tokens, numTokens ); // must free all mallocs
return 0;
}
Original phrase = "A rolling stone gathers no moss"
Number of words is 6
look at 2nd word one char at a time
r o l l i n g
Word List:
0: len= 1: A
1: len= 7: rolling
2: len= 5: stone
3: len= 7: gathers
4: len= 2: no
5: len= 4: moss
I recommend that you use a graphical debugger such as ddd to be able to view the memory layouts of your dynamic arrays.
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char **spliced;
use ...
char spliced[20][20];
(Or whatever dimensions that you need)