Copying Command Line Arguments Into an Array
For a program, I would like to make an array copy of the arguments sent in by command line using malloc().
So for example if I do ./a.out one two three I want an array with {a.out, one, two, three} in it.
However, I have some issues getting my program to work. Here's what I have:
static char** duplicateArgv(int argc, char **argv)
{
char *array;
int j = 0; 开发者_运维问答
// First allocate overall array with each element of char*
array = malloc(sizeof(char*) * argc);
int i;
// For each element allocate the amount of space for the number of chars in each argument
for(i = 1; i < (argc + 1); i++){
array[i] = malloc(strlen(*(argv + i)) * sizeof(char));
int j;
// Cycle through all the chars and copy them in one by one
for(j = 0; j < strlen(*(argv + i)); j++){
array[i][j] = *(argv + i)[j];
}
}
return array;
}
As you might imagine, this doesn't work. I apologize ahead of time if this somehow totally doesn't make sense, as I just started learning about pointers. Also, I'm not quite sure how to write code to free up every element in the *array after I do what I need to the copy.
Could anyone give me some tips on what I should look into to make it do what I want?
Thanks for any help!
You're not allocating or copying the terminating NULL characters:
This line needs to be changed to this for the NULL.
array[i] = malloc((strlen(*(argv + i)) + 1) * sizeof(char));
And the loop should be changed to this:
for(j = 0; j <= strlen(*(argv + i)); j++){
Also, the code can be better optimized if you saved the result of the strlen()
call since you call it in so many places.
Try the loop as this:
// For each element allocate the amount of space for the number of chars in each argument
for(i = 0; i < argc; i++){
int length = strlen(argv[i]);
array[i] = malloc((length + 1) * sizeof(char));
int j;
// Cycle through all the chars and copy them in one by one
for(j = 0; j <= length; j++){
array[i][j] = argv[i][j];
}
}
first you need to allocate a vector of char*, not just a char*
char **array;
array = malloc(sizeof(char*)*(argc+1)); // plus one extra which will mark the end of the array
now you have an array[0..argc]
of char*
pointers
then for each argument you need to allocate space for the string
int index;
for (index = 0; index < argc; ++index)
{
arrray[index] = malloc( strlen(*argv)+1 ); // add one for the \0
strcpy(array[index], *argv);
++argv;
}
array[index] = NULL; /* end of array so later you can do while (array[i++]!=NULL) {...} */
With
char *array;
you define an object of type char*
. That is: an object which value can point to a char
(and the next char
, ..., ...)
You need
char **array;
With this new type, the value of array
points to a char*
, ie another pointer. You can allocate memory and save the address of that allocated memory in a char*
, you't do that with a char
.
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