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copied!<p>My analysis below. From what I see here you should brush up on the difference between a pointer to char (char*) and a char array. I really appreciate though that you've tried to solve it yourself before asking.</p> <pre><code>const char *sCopy(char buffer[256], int i); /* let's start from here, what i represents? Keep in mind that the most of the */ /* time an external developer will see just your declaration of a method, always */ /* try to give significant names to variables. */ int main() { int i = 0; /* always initialize variables to default values, especially if they are */ /* going to be indexes in a buffer. */ int x = 0; char buffer[256] = ""; /* you can even initialize this to "" in order to mimic an empty string, */ /* that is a char array cointining just \0 (string null-terminator). */ char newBuffer[256] = ""; /* same here, you always need to declare the size of a char array unless */ /* you initialize it like this -char newBuffer[] = "hello"-, in which case */ /* the size will automatically be 6 (I'll let you discover/understand */ /* why 6 and not 5). */ printf("Please enter a number: "); fgets(buffer, 256, stdin); // nice link at the bottom on input reading i = atoi(buffer); printf("The value you entered is %d. Its double is %d.\n", i, i*2); newBuffer = sCopy(buffer, i); printf(newBuffer); return 0; } /* I am not going to debate the idea of returning a char pointer here :) */ /* Remember that in this case you are returning a pointer to some memory that has */ /* been allocated somewhere inside your function and needs to be released (freed) */ /* by someone outside your control. Are they going to remember it? Are they */ /* going to do it? In this case "they" is "you" of course. */ /* I'll let you explore alternative approaches. */ const char *sCopy(char buffer[256], int i){ char nBuffer[256] = ""; // see above char *t = NULL; /* you always init a pointer to NULL. As you can see, initializing here will */ /* make you think that there might be problem with the strcat below. */ int x; // ok here you can omit the init, but be aware of it. for(x = 0; x < i; x++){ strcat(t, buffer); /* what are you missing here? this piece of code is supposed to concatenate the */ /* input buffer to a brand new buffer, pointed by your variable t. In your implementation */ /* t is just a pointer, which is nothing more than a number to a memory location */ /* With the initialization, the memory location you are pointing to is NULL, which */ /* if de-referenced, will cause massive problems. */ /* What you are missing is the blank slate where to write your data, to which your */ /* pointer will read from. */ } //t = nBuffer; return t; } </code></pre> <p>I really hope that this would help you. I am sorry but I can't write the solution just because I think it's better if you learn it the hard way. You can find plenty of tutorials of pointers to char and I am sure you will solve the problem.</p> <p>(input reading) <a href="https://stackoverflow.com/questions/7060573/c-scanf-and-fgets-problem">C scanf() and fgets() problem</a></p>

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