I am just learning C and making a basic "hello, NAME" program. I have got it working to read the user's input but it is output as numbers and not what they enter?
What am I doing wrong?
#include <stdio.h>
int main()
{
char name[20];
printf("Hello. What's your name?\n");
scanf("%d", &name);
printf("Hi there, %d", name);
getchar();
return 0;
}
You use the wrong format specifier %d- you should use %s. Better still use fgets - scanf is not buffer safe.
Go through the documentations it should not be that difficult:
scanf and fgets
Sample code:
#include <stdio.h>
int main(void)
{
char name[20];
printf("Hello. What's your name?\n");
//scanf("%s", &name); - deprecated
fgets(name,20,stdin);
printf("Hi there, %s", name);
return 0;
}
Input:
The Name is Stackoverflow
Output:
Hello. What's your name?
Hi there, The Name is Stackov
#include <stdio.h>
int main()
{
char name[20];
printf("Hello. What's your name?\n");
scanf("%s", name);
printf("Hi there, %s", name);
getchar();
return 0;
}
When we take the input as a string from the user, %s is used. And the address is given where the string to be stored.
scanf("%s",name);
printf("%s",name);
hear name give you the base address of array name. The value of name and &name would be equal but there is very much difference between them. name gives the base address of array and if you will calculate name+1 it will give you next address i.e. address of name[1] but if you perform &name+1, it will be next address to the whole array.
change your code to:
int main()
{
char name[20];
printf("Hello. What's your name?\n");
scanf("%s", &name);
printf("Hi there, %s", name);
getchar();
getch(); //To wait until you press a key and then exit the application
return 0;
}
This is because, %d is used for integer datatypes and %s and %c are used for string and character types
Related
int main()
{
int Age;
char Name;
//Age
printf("Type your age: ");
scanf_s("%d", &Age);
printf("Your age is %d\n", Age);
//Name
printf("Type your Name: ");
scanf_s("%s", &Name);
printf("Your name is %s", Name);
return 0; }
It's the 'Name' section which is throwing out an error. I can't figure out why.
UPDATE: I'm coding in Visual Studio. Therefore, "scanf_s" is essentially required.
The error is "Exception thrown at 0x5B49D4EC (ucrtbased.dll) in Project1.exe: 0xC0000005: Access violation writing location 0x001A0000. occurred"
Your problem is that char Name; can only store a single character. Your code is allowing the user to type in multiple characters which are being stored into Name causing a memory error.
Change char Name; to something like char Name[50] so that you can store up-to 49 characters plus the null byte.
Also you should use scanf_s() properly to avoid the error if the buffer (char array) ends up being too small.
Note, you should always check the return from scanf_s() so you know if the user entered valid data or not.
This code works correctly in Visual Studio:
#include "stdafx.h"
#include <string.h>
#include <stdlib.h>
int main()
{
int Age;
char Name[50];
printf("Type your age: ");
if(scanf_s("%d", &Age))
{
printf("Your age is %d\n", Age);
printf("Type your Name: ");
if (scanf_s("%s", Name, (unsigned)_countof(Name)))
{
printf("Your name is %s\n", Name);
}
else
{
printf("Name:: Invalid Input\n");
}
}
else
{
printf("Age:: Invalid Input\n");
}
return 0;
}
The problem is that you defined Name as a char - a single character - but you are trying to use it as a string (multiple characters).
To fix this you must either (a) define Name as an array of characters (which would be a string) - such as char Name[100]; or (b) as a pointer (such as char *Name;) - which would require you to malloc() the string before use and free() it after use.
Strings can be tricky, as they are basically just arrays of chars, but that requires you to either know, or find a way to know, how many characters will be in the string. You can read more about how to do that here, in the documentation for scanf_s, which gives this example:
char c[4];
scanf_s("%4c", &c, (unsigned)_countof(c)); // not null terminated
First off I would just use scanf(), not scanf_s().
Furthermore you need to cast your Name variable as a string, which is an array of characters as I have defined it below. Using just char Name, means you have created a variable with room for just one character.
Hope this helps :)
int main()
{
int Age;
char Name[10];
printf("Type your age: ");
scanf("%d", &Age);
printf("Your age is %d\n", Age);
//Name
printf("Type your Name: ");
scanf("%s", &Name);
printf("Your name is %s", Name);
return 0;
}
Fixed the problem by going to...
Tools->Options->Debugging->Symbols and select checkbox "Microsoft Symbol Servers", Visual Studio will download PDBs automatically.
Thanks for everyone's help :)
When I give the input then only first alphabet is showing.
I want to print the complete name which is I just entered.
#include <stdio.h>
int main()
{
char name;
char grades;
int i;
printf("Name of the Student:");
scanf("%c",&name);
printf("Name your Just entered is : %c",name);
return 0;
}
I agree with the others - but add some error checking and ensure no buffer overruns i.e
#include <stdio.h>
int main() {
char name[101];
printf("Name of the student:");
if (scanf("%100s", &name) == 1) {
printf("Name you just entered: %s\n", name);
return 0;
} else {
printf("Unable to read name of student\n";
return -1;
}
}
EDIT
As you have edited the question so that it does not have the same meaning as before I will leave my previous solution here.
But what you want is to use fgets - this allows for white space in the name
ie.
#include <stdio.h>
int main()
{
char name[100];
printf("Name of student:");
fflush(stdout);
fgets(name, 100, stdin);
printf("Students name is %s\n", name);
return 0;
}
Replace char name; with char name[100];. This will define name as array of chars, because you handled with it as single character.
For scanf replace it with scanf("%s",&name[0]);, and printf with printf("Name your Just entered is : %s",name);. %s means string, so it will scan whole string, not just single character. In scanf &name[0] points to beginning of array.
You need to scanf into an array, rather than into a single character:
#include <stdio.h>
int main() {
char name[100];
printf("Name of the student:");
scanf("%s", &name);
printf("Name you just entered: %s\n", name);
}
You are trying to store a array of characters(string) in a character. So only the first character is taken.To rectify this initialize the name as:
char name[40];
take input as :
scanf("%s",name);
and print as:
printf("name is %s",name);
name is a char and scanf will only catch one character when you use %c. You can use a char array to store the name instead :
char name[40];
/* edit the size for your need */
Also edit your scanf and printf to use a %s
You are reading (and printing) a single char using %c. If you want to handle stirngs, you should use a char[] and handle it with %s:
#include <stdio.h>
int main()
{
char name[100]; /* Assume a name is no longer than 100 chars */
char grades;
int i;
printf("Name of the Student: ");
scanf("%s",&name);
printf("Name your Just entered is : %s",name);
return 0;
}
#include "stdafx.h"
int main() {
char name;
printf("What is your name:"); // I enter my name..
scanf_s("%c", &name); // Should grab my name in this case (Brian)
printf("Hello, %c\n", name); //Should print "Hello, Brian."
return 0;
}
What's wrong? Why is it not storing the whole name and just the first letter?
Because name is only a single char and you've used %c in scanf (and printf()).
Looks like what you want is:
char name[32]; /* assumes the name is 31 chars or less */
printf("What is your name:"); // I enter my name..
scanf_s("%s", &name); // Should grab my name in this case (Brian)
printf("Hello, %s\n", name); //Should print "Hello, Brian."
But since you have flagged the question as C++, then there are better ways of doing it than this!
Missing parameter:
The fscanf_s function is equivalent to fscanf except that the c, s, and [ conversion specifiers apply to a pair of arguments ... That argument is immediately followed in the argument list by the second argument, which has type rsize_t and gives the number of elements in the array pointed to by the first argument of the pair. C11 §K.3.5.3.2 4
// scanf_s("%c", &name);
scanf_s("%c", &name, (rsize_t) 1);
This fixes the code's undefined behavior, but still will only save/print 1 char of input.
Code should read in a number of char as a string.
#define NAME_SIZE 100
char name[NAME_SIZE];
scanf_s("%s", &name, (rsize_t) NAME_SIZE);
printf("Hello, %s\n", name);
This question already has answers here:
Why is getchar() reading '\n' after a printf statement?
(3 answers)
Closed 9 years ago.
This is a newbie question. I am new to C programming. I have the following code which does not prompt for 'Name' Onece the 'Age' is entered, it bypass the 'Name section.
#include <stdio.h>
int main()
{
char name[30],ch;
int age;
printf("Enter age : ");
scanf("%d", &age);
int i=0;
printf("Enter name: ");
while((ch = getchar())!='\n')
{
name[i]=ch;
i++;
}
name[i]='\0';
printf("Name: %s\n",name);
printf("Age : %d\n", age);
return 0;
}
After reading first prompt it bypass the second prompt which is using getchar() function. But if I change the order of prompt to ask for 'Name' first and then 'Age' it works fine.
The working code.
#include <stdio.h>
int main()
{
char name[30],ch;
int age;
int i=0;
printf("Enter name: ");
while((ch = getchar())!='\n')
{
name[i]=ch;
i++;
}
name[i]='\0';
printf("Enter age : ");
scanf("%d", &age);
printf("Name: %s\n",name);
printf("Age : %d\n", age);
return 0;
}
My coding IDE is CodeBlock and my compiler is GNU C Compiler (mingw32-gcc.exe)
Please help me to breakthrough.
A few improvements/advices to the code in the question:
the type of the return value of getchar() is int, so the type of ch also should be int
you could (and should, I believe) use format %s to read the name, this is easier and the leading white spaces in the input stream would not be a problem
the user of the code could give a name which contains more than 30 characters, and this input could crash your program, so you should protect your code for this possibility. You have two options:
a. use format '%29s" to read the name
b. change the definition of name to char *name, read it by scanf("%ms", &name);, and call free(name); after you do not need it anymore
Here is an example, in which the name can be very long and can include spaces:
#include <stdio.h>
#include <stdlib.h>
int
main(int argc, char *argv[])
{
char *name;
int age;
printf("Enter name: ");
scanf("%m[^\n]", &name);
printf("Enter age: ");
scanf("%d", &age);
printf("Name: %s\n", name);
printf("Age : %d\n", age);
free(name);
exit(EXIT_SUCCESS);
}
And here is a run of it:
$ ./a.out
Enter name: a very looooooooooooooooooooooooooooooooooooooooooooooong name
Enter age: 12
Name: a very looooooooooooooooooooooooooooooooooooooooooooooong name
Age : 12
In first code the \n character left behind by the scanf is read by getchar. This makes the condition (ch = getchar())!='\n' in while loop false and the loop body never get executed.
You need to consume that \n character which comes up to the buffer along with the age you entered on pressing Enter key.
Putting the statement
while(getchar()!='\n');
after the scanf will consume all of the newline characters.
Your second code is working fine because %d skips white-space characters unlike %c specifiers.
So I want to make the hello.c program write both the first name and the last name on one line so in this form but when I run my program in this current form it gives me the error "expected â)â before string constant" I think I have the rest of the code down because I have removed that line and ran it and it works. So I just want to ask how to get 2 strings that I have already pointed to to go on the same line.
This is my code
#include <stdio.h>
int main()
{
char firstname[20];
char lastname[20];
printf("What is your firstname?");
scanf("%s", firstname);
printf("What is your lastname?");
scanf("%s", lastname);
printf("Hello %s\n", firstname "%s", lastname);
printf("Welcome to CMPUT 201!");
}
You want
printf("Hello %s %s\n", firstname, lastname);
instead of
printf("Hello %s\n", firstname "%s", lastname);
#include<stdio.h>
#include<string.h>
int main()
{
char first_name[20] = " ", last_name[20] = " ", full_name[40] = " ";
printf("What is your firstname?\n");
scanf("%s", first_name);
printf("What is your lastname?\n");
scanf("%s", last_name);
sprintf(full_name,"%s %s",first_name, last_name);
printf("name is %s\n",full_name);
return 0;
}
I have shown the same using sprintf.
1) also in your program you are not returning anything, if dont want to return anything make it as void function. When you write int function, always make it an habbit to return the integer.
2) Also when you write the printf function always make it habit to add \n(new line) so the output looks good
happy coding.