There are several problems with your code: You’re error checking isn’t correct. You check if(argc!=2||!apha) after you’ve already evaluated strlen(argv[1]) — by then it’s too late! Check the validity of argc before accessing argv and don’t double up the argument count error and alphabetic key error, they’re independent. Also, error messages should go to stderr, … Read more
You may be looking for the declaration of additional blocks as in #include<stdio.h> int main() { int i=1; printf(“before %d\n”,i); { int i=0; printf(“within %d\n”,i); } printf(“after %d\n”,i); return(0); } which when executed yields before 1 within 0 after 1 From the comment I grasp that a function invocation may be preferable which may be … Read more
Based on the phrasing of your question, I’m guessing that you are unfamiliar with the concept of pointers. In C, pointer is a type that stores the memory address of another variable. The & operator retrieves the memory address of a variable. int i = 4; // < An integer variable equal to 4 int* … Read more
sprintf formats a string and writes it into the character array specified by the first argument (assuming sufficient space); printf formats a string and writes it to stdout. Ex: you can do this with sprintf: char buffer[100]; sprintf(buffer, “My name is %s and I am %d years old”, “John Doe”, 25); // buffer now contains … Read more
A function to caesar an alphabetic char should be like (decomposed in elementary steps): int caesar_lower(int c,int key) { int v = c-‘a’; // translate ‘a’–‘z’ to 0–25 v = v+key; // translate 0–25 to key–key+25 v = v%26; // translate key–key+25 to key–25,0–key-1 v = v+’a’; // translate back 0–25 to ‘a’–‘z’ return v; … Read more
Actually, the function returns a value. Maybe you want to see it, so just print the result: #include <stdio.h> #include <cs50.h> int calcrectarea(int len,int wid){ return len*wid; } int main(){ printf(“enter length here:\n”); int x; scanf(“%d”,&x); printf(“enter width here:\n”); int y; scanf(“%d”, &y); printf(“area is: %d”, calcrectarea(x,y)); } solved The function is not returning in … Read more
Ok, you dynamically allocate 10 bytes. Then you make the “name” variable (a char pointer) point to that dynamically allocated block. Then you make a char pointer which points to the same dynamically allocated block that “name” points to. Then you “free(name)” which deallocates the block that “tmp_name” also points to. “name” and “tmp_name” both … Read more