Problems passing in a char[][] to execvp - c

Question

I am trying to iterate over some input (which are commands and arguments), split the input into individual strings, then pass that input into execvp().
I am having trouble as execvp() wants a (char*, char*[]) as its arguments. I am passing in (char*, char[][]) which I thought was the same thing but it isn't liking it.
I would use a char*[] but I don't know how big the strings are prior to it running, so that is the reason I didn't use it. So obviously if I use char*[], I get a seg fault when I try and access elements of the char*'s.
Here's a snippet of the code
//input
char *line = "echo hello there what is";
int count = 0;
//store arguments
char args[6][10];
int argCount = 0;
//store temp string from input
char temp[100];
int tempCount = 0;
//word(string) size
int wordSize = 0;
/*
Here I iterate over the input, storing each string I find in args
all the above variables help me do that.
*/
execvp(args[0], args);
printf("Error: It didnt work\n");
Hopefully that is clear and a valid question, let me know if you want me to add the code of me turning the input into args.

Answer 1

An array of arrays of char is not the same as an array of pointers to arrays of char. execvp() expects the latter, with a null pointer as its last element, passing the former has undefined behavior.
You must construct an array of pointers, either allocated from the heap or defined with automatic storage (on the stack), initialize it with pointers to argument strings and pass this array to execvp().
Note also that echo is both a shell internal command and an executable file in the path.
Here is your code fragment modified accordingly (without the parse code, which is still yours to write):
//input "echo hello there what is";
//arguments array
char *args[6];
/*
* Here you should iterate over the input, storing each string you find
* on the command line into `args` and terminate with a null pointer...
*/
args[0] = "echo";
args[1] = "hello";
args[2] = "there";
args[3] = "what";
args[4] = "is";
args[5] = NULL;
execvp(args[0], args);

Answer 2

You can use two arrays:
char real_args[6][10]; // Six strings of up to nine characters each
...
char *args[] = {
real_args[0],
real_args[1],
real_args[2],
real_args[3],
real_args[4],
real_args[5],
NULL // Must be terminated by a null pointer
};
Use real_args for the actual arguments, and then pass args to execvp.

Answer 3

Ok I worked out how to use a char*[] instead of a char[][] and successfully put it into execvp().
Notes for the code below: I keep track of how long the current string is that I am iterating over in wordSize.
//input
char* line = "echo hello there what is";
int count = 0;
//store arguments
char* args[10];
int argCount = 0;
//store temp string from input
char temp[100];
int tempCount = 0;
//word size
int wordSize = 0;
while(line[count] != '\0')
{
if (line[count] == ' ' || line[count + 1] == '\0')
{
/*
As I don't know how big each string will be, I can
allocate it here as I know how big the string is through wordSize
*/
args[argCount] = malloc(sizeof(char)*(wordSize +1));
.
.
.
}
}
//terminate args with 0
args[argCount] = 0;
execvp(args[0], args);
printf("Error: It didnt work\n");