Why following code snippets assignment gives confusing output? - c

Question

I'm studying C. I came across with string arrays. I'm bit confused about the following codes. I was anticipating one kind of output; however, getting completely different kind of output or program crush due to read access violation.
I've run this code on visual studio 2017, with _CRT_SECURE_NO_WARNINGS
// case 1
char* name[2];
//name[0] = (char*)malloc(sizeof(char*) * 10);
//name[1] = (char*)malloc(sizeof(char*) * 10);
name[0] = "john";
name[1] = 'doe';
printf("%s\n", name[0]); // prints john
//printf("%s\n", name[1]); // gives read access violation exception, why??? even with dynamically allocated memory
// case 2
char* name2[2] = { "emma", "olsson" };
printf("%s\n", name2[0]); // prints emma
printf("%s\n", name2[1]); // prints olsson, why no error???
// case 3
for (int i = 0; i < 2; i++)
{
name[i] = name2[i];
}
printf("%s\n", name[0]); // prints emma
printf("%s\n", name[1]); // prints olsson, why no error???
// case 4
char inputName[10];
int i = 0;
while (i < 2)
{
fgets(inputName, sizeof(inputName), stdin); // first input: Max second input: Payne
char* pos = strchr(inputName, '\n');
if (pos != NULL)
*pos = '\0';
name[i++] = inputName;
}
printf("%s\n", name[0]); // prints Payne, why not Max???
printf("%s\n", name[1]); // prints Payne

Answer 1

For case 1, 'doe' is not a string.
Case 2 works because you are initializing you pointers with string literals.
Case 3 works too because you assign the same initialized pointer in case 2 to case 1 pointers. Your name array pointers are basically set to point to where name2 ones are pointing.
In case 4, you declared inputName which points to a set of 10 chars. Then each time you get a new input you are writing it to the same memory section. And by doing this:name[i++] = inputName;
you are not copying a new char array to name[i] as you might think. Instead, you are telling name[i] char pointer to point to inputName. So it is normal that name prints last input twice, because that's what inputName points to, as well as both name char pointers.

Answer 2

It is unclear whether OP's code runs within main() or a user-defined function and what kind of value returns. That said, after removing superfluous variable redeclarations, here's how I achieved working code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
char * name[2];
char * name2[2]={ "emma", "olsson" };
char inputName[10];
char names[BUFSIZ];
int i = 0;
// case 1
name[0] = "john";
name[1] = "doe";
printf("%s %s\n", name[0],name[1]); //john doe
// case 2
printf("%s %s\n", name2[0],name2[1]);//emma olsson
// case 3
for (i = 0; i < 2; i++){
name[i] = name2[i];
}
printf("%s %s\n", name[0],name[1]);//emma olsson
// case 4
i=0;
while (fgets(inputName, sizeof(inputName), stdin) != NULL && (i < 2) ){
strcat(names,inputName);
i++;
}
printf("\n%s\n",names);
return 0;
}
See live code here
OP should replace the single quotes around doe with double quotes which denote a null-terminated string. Single quotes are meant for a single character, i.e. 'a' refers to a byte value while "a" signifies a string containing two characters, an 'a' and a '\0'.
Also, OP should include two other libraries to facilitate execution. In particular, string.h is needed for the built-in string functions to execute properly.
Case 2 and Case 3 work because strings are encompassed by double quotes instead of single quotes. Note that in each case the "%s" format specifier for the printf() indicates that a string needs to be displayed.
In last case, fgets() with respect to stdin, upon success returns user input as a string. But that input will be overridden in the while loop, unless in each iteration you concatenate the old input with the new. Otherwise, when the inputName element values change because its address remains constant, only the latest input string displays. Here's some code that illustrates this point:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
char * name[2];
char inputName[10];
int i = 0;
while (fgets(inputName, sizeof(inputName), stdin) != NULL && (i < 2) ){
printf("inputName: %p points to: %s",inputName,inputName);
name[i++] = inputName;
}
printf("\n name[0]: %p points to: %s\n name[1]: %p points to: %s",name[0],name[0],name[1],name[1]);
return 0;
}
Output:
inputName: 0x7fff8a511a50 points to: Maxine
inputName: 0x7fff8a511a50 points to: Pauline
name[0]: 0x7fff8a511a50 points to: Pauline
name[1]: 0x7fff8a511a50 points to: Pauline
See live code.
Incidentally, you don't need an array to display the names and indeed one may display the names outside of the loop as long as within the loop the code concatenates user input.