Dynamic memory allocation for pointer arrays

Question

I'm am trying to write a program that reads in a series of strings from a text file and stores these in an array of strings, dynamically allocating memory for each element. My plan was to store each string in an array using a pointer and then grow the array size as more were read in. I am having trouble to understand why my test code below is not working. Is this a workable idea?

char *aPtr; aPtr =(char*)malloc(sizeof(char));  aPtr[0]="This is a test";   printf("%s",aPtr[0]); 

Answer 1

In C a string is a char*. A dynamic array of type T is represented as a pointer to T, so for char* that would be char**, not simply a char* the way you declared it.

The compiler, no doubt, has issued some warnings about it. Pay attention to these warnings, very often they help you understand what to do.

Here is how you can start your testing:

char **aPtr; int len = 1; // Start with 1 string aPtr = malloc(sizeof(char*) * len); // Do not cast malloc in C aPtr[0] = "This is a test"; printf("%s",aPtr[0]); // This should work now. 

Answer 2

char *str; //single pointer    

With this you can store one string.

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To store array of strings you Need two dimensional character array

or else array of character pointers or else double pointer

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char str[10][50]; //two dimensional character array 

If you declare like this you need not allocate memory as this is static declaration

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char *str[10];  //array of pointers  

Here you need to allocate memory for each pointer

loop through array to allocate memory for each pointer

for(i=0;i<10;i++)  str[i]=malloc(SIZE); 

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char **str;    //double pointer 

Here you need to allocate memory for Number of pointers and then allocate memory for each pointer .

str=malloc( sizeof(char *)*10); 

And then loop through array allocate memory for each pointer

for(i=0;i<10;i++)  str[i]=malloc(SIZE);