HELP!!!

Question

char *string[]={“dog”,”ant”,”cat”,”elephant”};
instead of using declaration like above, may i know how can i get user input for strings of character?
Thanks for helping!

Answer 1

Use one for loop and in inner part of loop write scanf and use %s

Answer 2

for (int i=0; i<4; i++) scanf("%s", &string[i]);

Answer 3

Here.
https://code.sololearn.com/cr7ZLPsyHJWK/?ref=app

Answer 4

Jia Yin &string[i] is wrong. Try scanf("%s", string[i]);

Answer 5

You need a multidimension char array to store user inputs. char pointer (char*) are constant and non write-able.

#include <stdio.h>

int main()
{
char animals[4][20]; // 4 rows @ 20 char each
for(size_t i = 0; i < 4; i++)
{
printf("Enter animal %lu > ", i + 1);
fflush(stdout);
// read up to 19 char. Leave the last
// character as string terminator.
scanf("%19s", animals[i]);
}
for(size_t i = 0; i < 4; i++)
{
printf("Animal %lu > %s\n", i + 1, animals[i]);
}
return 0;
}

Answer 6

SapphireBlue you are right... sorry... 🙄

Answer 7

c or c++?

Answer 8

Its C language

Answer 9

for (int i=0; i<4; i++) scanf("%s", &string[i]);
I tried, but it wasnt work.

Answer 10

Jia Yin

You cannot do that for char *string[] = { . . . } since that's an array of pointers to read-only memory. You need a sized buffer to store the characters from stdin.

char string[4][32];

for (int i = 0; i < 4; i++)
scanf("%31s", string[i]);

I would also recommend using fgets() over scanf(). The scanf() functions have a whole range of issues that make them a poor choice for user input. If you're going to use it, at least make sure it doesn't go past the string buffer by defining the width in the format specifier.

Answer 11

pls help.