如何在 C 中创建字符串类型的变量

Question

Question

How to declare a string variable in C?

Background

In my quest to learn the basics of c , I am trying to port one of my oldest python programs, Bob , to C. In the program, the script asks the user for information on him or herself, and then spits out responses. Almost all of these variables use raw_input for their information - the variables are strings. But, I have found no way to declare C variables.

Code

So far, I have tried to declare the variable as of type char and int. Here is the code, switch the type at your leisure.

int main(int argc, const char * argv[])
{

    int name;
    printf("What is your name?");
    scanf("%s",&name);
    printf("Your name is %s", name );

    return 0;
}

Error Message

When I run this code, Xcode returns some weird stuff. This part of the globidty-gloop is highlighted.

0x7fff96d2b4f0:  pcmpeqb(%rdi), %xmm0

Lasty, this Yahoo Answer said that I had to use something called a character array . It was posted 5 years ago, so I assumed that there was a better way.

EDIT

I am following the tutorial at C Programming .

Answer 1

char name[60];
scanf("%s", name);

Edit: restricted input length to 59 characters (plus terminating 0):

char name[60];
scanf("%59s", name);

Answer 2

The int your putting is not a string, a string looks like "char myString[20]". Not like "int name", that's an integer and not a string or char. This is the code you want:

         int main(int argc, const char * argv[])
{

char name[9999];
printf("What is your name?\n");
scanf("%s", name);
system("cls");
printf("Your name is %s", name);

return 0;
}

Answer 3

In C you can not direct declare a string variable like Java and other language. you'll have to use character array or pointer for declaring strings.

char a[50];
printf("Enter your string");
gets(a);

OR

char *a;
printf("Enter your string here");
gets(a);

OR

char a[60];
scanf("%59s",a);

Answer 4

TESTED ON XCODE

You can do so:

int main(int argc, const char * argv[])
{

    int i;
    char name[60]; //array, every cell contains a character

    //But here initialize your array

    printf("What is your name?\n");
    fgets(name, sizeof(name), stdin);
    printf("Your name is %s", name );

    return 0;
}

Initialize the array, is good to avoid bug

for(i=0;i<60;i++){
      name[i]='\0'; //null
}

Instead int is used for int number (1, 2, 3, ecc.); For floating point number instead you have to use float

Answer 5

C does not have a string variable type. Strings can be stored as character arrays (char variable type). The most basic example I would add up to the rest is:

int main()
{
   char name[] = "Hello World!";
   printf("%s",name);
   return(0);
}

Answer 6

It's easy!

Just put this line below, atop of your main() function.

typedef string char*;

This allows you to create a string variable as you do with integers or characters in C. After that, your program should look like this:

#include <stdio.h>

typedef char* string;

int main(void) {
    string a = "Hello";
    printf("%s\n", a);  // %s format specifier for String
    return 0;
}

For a live demonstration , visit this REPL.it .

Answer 7

Normally we use "&" in scanf but you shouldn't use it before variable "name" here. Because "name" is a char array. When the name of a char array is used without "[]", it means the address of the array.

Answer 8

replace int name; to--. char name[60];

#include <stdio.h>
int main()
{
  char name[648];
  printf("What is your name?");

  scanf("%s", name);
  printf("Your name is %s", name );

  return 0;
}