How to get the value of a string from char* not the address

Question

I am trying to compare patterns. So I have structs which hold the patterns as strings, however I want to be able to build a string and store the VALUE of that string in the struct. At the moment, I am only copying the address of my string.

typedef struct
{
    int emp;
    char *context;
    int numOnes;
    int numZeros;


}Pattern;

    char *patt, str[t+1];
    patt = str;
    while(count<t){
        //printf("Stream: %d\n", stream[end]);
        if(stream[end] == 1){
            patt[count]= '1';
            //writeBit(1);
        }
        else{
            patt[count]='0';
            //writeBit(0);
        }


        end--;
        count++;
    }
    patt[count]=0;//building string


    if(found == 0){//if pattern doesnt exist, add it in

        patterns[patternsIndex].context = patt; //NEED HELP HERE. This copies the address not the actual value, which is what i need

        patterns[patternsIndex].emp = 1; 
        prediction = 0;
        checkPredict(prediction,stream[end],patternsIndex);
        patternsIndex++;

        found =1;
} 

Answer 1

To avoid making Pattern.context a fixed size array and to be certain there is enough space to copy into you will need to dynamically allocate memory for Pattern.context to store a copy of patt. Either:

• use malloc() (remembering to allocate strlen(patt) + 1 for the null terminator) and strcpy():

  patterns[patternsIndex].context = malloc(strlen(patt) + 1);
  if (patterns[patternsIndex].context)
  {
      strcpy(patterns[patternsIndex].context, patt);
  }
  

• use strdup():

  patterns[patternsIndex].context = strdup(patt);
  

In either case, remember to free() the string copy when no longer required:

free(patterns[patternsIndex].context);

Answer 2

If you mean you want to take the string "14" and make it be a number with the value of 14, then look at the atoi standard function.