How to create a string-type variable in C

问题:

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.

回答1:

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

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

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

回答2:

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);

回答3:

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;
}

回答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

回答5:

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.

回答6:

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;
}