Newbie here. C function problem during execution - c

Question

Edit: Since I understand that I need to provide more info to make it clear for you guys, I added the main function and the getchoice and also two images of the program running. My problem is that after entering the endword, I want to see the menu first and then make a choice, whereas it prompts me to give an input without showing the menu.
This function is part of a bigger program, but this is where a problem occurs.
It reads words inputed, places them into an array, until the keyword ****END is entered. However, when this keyword is entered, it doesn't go immediatelly in the specified if clause (you will see that in the code). I'm a newbie and it could be something really obvious, but any help is greatly appreciated.
#include <string.h>
#define M 50
#define N 15
void getText(char a[M][N])
{
int i, j;
char temp[N];
for (i = 0; i < 50; i++) {
for (j = 0; j < 15; j++) {
if (i == 49 && j == 14) {
printf("Maximum length of text reached.\n");
}
scanf("%s\n", temp);
if (strcmp(temp, "****END") == 0) {
printf("You entered the endkey.\n");
return;
}
strcpy(a[i], temp);
}
}
}
int main(){
int input;
while(1){
input = getChoice();
if(input == 1){
getText(text);
}
else if(input == 2){
getDictionary();
}
else if(input == 3){
correctText();
}
else if(input == 4){
saveText();
}
else if(input == 5){
getStats();
}
else if(input == 6){
break;
}
}
return 0;
}
int getChoice(){
int temp;
printf("Choose function:\n1: Enter text\n2: Enter dictionary\n3: Correct text\n4: Save text\n5: Get text statistics\n6: Exit program\n");
scanf("%d", &temp);
return temp;
}
Entered the endword and now it waits for input instead of showing the menu.
I inputed 2 for the second program function, then it showed the menu and proceeded to function 2.

Answer 1

Apart from the unnecessary double-nested loop, this line
scanf("%s\n", temp);
should be
scanf("%s", temp);
Usually, you should not try to match trailing whitespace with scanf, and the format specifier %s automatically filters out leading whitespace (but note that %c does not).
There are other faults and the code presented was originally incomplete, but notably the input length for %s must be restricted to prevent buffer overflow.

Answer 2

#include <stddef.h> // size_t
#include <ctype.h> // isspace()
#include <stdio.h> // scanf(), puts()
#include <string.h> // strcmp()
// see https://stackoverflow.com/questions/2653214/stringification-of-a-macro-value
#define STRINGIFY(x) #x
#define STRING(x) STRINGIFY(x)
#define LINES 50
#define COLS 15
char const *end = "****END";
// throw away everything until a newline is found
void clear(FILE *stream)
{
int ch;
while ((ch = getc(stream)) != EOF && ch != '\n');
}
size_t getText(char dst[LINES][COLS + 1])
{
size_t i = 0;
for (; i < LINES; i++) {
char temp[COLS + 1] = { 0 };
scanf("%" STRING(COLS) "s", temp); // "%15s" at runtime.
int ch;
// if the next character is not whitespace ...
if ((ch = getchar()) != EOF && !isspace(ch)) {
puts("Warning: Input too long, was truncated!");
clear(stdin);
}
if (strcmp(temp, end) == 0) {
puts("You entered the endkey.");
return i;
}
strcpy(dst[i], temp);
}
return i;
}
int main(void)
{
// COLS + 1 ... we need space for the terminating newline character.
char foo[LINES][COLS + 1];
size_t n = getText(foo);
for (size_t i = 0; i < n; ++i)
puts(foo[i]);
}
The %s conversion specifier should never be used without specifying a width to limit the characters that get stored:
char foo[10];
scanf("%9s");