Is it possible to read a character and an integer in one time?
Like this:
char name[32];
int age;
printf("Give name and age: ");
scanf("%s%d%*c", name, &age);
It blocks all the time.
UPDATED: You only accept input but not printing any thing. See the below code ( working and tested )
#include <stdio.h>
int main(void) {
char name[32];
int age;
printf("Give name and age: ");
scanf("%31s%d", name, &age);
printf("Your name is %s and age is %d",name,age);
}
intput: Shaz 30
Output : Your name is Shaz and age is 30
char name[32];
int age;
printf("Give name and age: ");
scanf("%31s%d", name, &age);
If the input string is longer than 31 characters, the value read for age will be affected, this is where fgets comes in handy.
Do not read both the name and age on the same line input. Read the name first to force the user to enter a new line so you can handle the entire line as the input name.
Read fgets
It is one of the method
CODE
#include<stdio.h>
char name[32],age;
int main()
{
printf("Enter name and age: ");
scanf("%s%d", name, &age);
printf("\nNAME : %s AGE : %d",name,age);
}
OUTPUT Refer under the image
I have tested your code, there is no error. Maybe you should only add printf() to print the answer.You will find you can get the same answer with what you have printed.
EDIT: If input is Colin 23(the space is necessary), you should use printf("name is %31s, age is %d", name, age). Then you will get output Colin 23.
You can do it like this:
#include <stdio.h>
int main(void)
{
char name[32];
int age;
printf("Give name and age: ");
scanf("%s%d", name, &age);
printf("%s\n",name);
printf("%d",age);
}
Related
Can't figure out why my printf output won't print the int data.age or addr.zip correctly. They're defined as int, so %d should work...but it doesn't. I get gibberish numbers for the output. All other fields work perfectly.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct BillingAddress {
char *street[50];
char *city[25];
char *state[3];
int *zip[6];
}address;
address addr;
typedef struct ContactInfo {
char *name[30];
int *age[3];
char *phone[15];
struct BillingAddress PersonalData;
}personaldata;
personaldata data;
int main()
{
printf("Enter your full name: ");
scanf(" %49[^\n]", &data.name);
printf("Enter your age: ");
scanf("%d", &data.age);
printf("Enter your Phone Number (xxx) xxx-xxxx: ");
scanf(" %14[^\n]", &data.phone);
printf("Enter your street address: ");
scanf(" %49[^\n]", &addr.street);
printf("Enter your city: ");
scanf(" %24[^\n]", &addr.city);
printf("Enter your state abbreviation: ");
scanf("%s", &addr.state);
printf("Enter your zip code: ");
scanf("%d", &addr.zip);
printf("Personal Data: \n %s\n %d\n %s\n %s\n %s\n %s\n %d\n", data.name, data.age, data.phone, addr.street, addr.city, addr.state, addr.zip);
}
Let's go step by step. In your BillingAddress struct definition, you intend to store the street, city, and state. It seems logical that you could represent all of those parameters as strings (which are character arrays in C). I believe that is what you tried to do, but the declaration:
char *street[50];
represents give you an array of 50 char * (i.e., char pointers) NOT 50 chars, which is what you actually want. So, the struct definition should instead be:
typedef struct BillingAddress {
char street[50];
char city[25];
char state[3];
int zip;
} address;
You also intend to store the zip. Note how I changed the type of zip to an int array rather than an int array. You could do an int array, but you would have to do more work to properly accept the user's input (simply using scan("%d", &addr.zip) will not work).
Now, let's take a looked at a different way to define your definition for ContactInfo struct:
typedef struct ContactInfo {
char name[30];
int age;
char phone[15];
address addr;
} personaldata;
You can see that I've removed the *'s and changed age to be a single int for the similar reasoning as what I mentioned above. I've also replaced struct BillingAddress with the simpler address. Why? Because you created a typedef linking the keyword address to struct BillingAddress. So, by using address alone, the code looks a bit cleaner and you are putting that typedef to good use :D (Note that there's nothing incorrect about using struct BillingAddress, I just think it looks cleaner to utilize the typedef).
Also, you'll see that I renamed the variable of type address (i.e., of type struct BillingAddress -- remember the typedef means these two are the same) to addr. This makes more sense than naming it PersonalData as you did, because well, it's an address, not arbitrary personal data.
Now moving onto your main() method. You should be aware of some of the pitfalls of using scanf(). Since it reads from the default input stream stdin (i.e., most likely your terminal), any characters not gulped up by scanf() will be left in the input stream. So, any subsequent calls to scanf() will also read in those unwanted characters. See this post for more info. You might not have run into this if you supplied the less than the max number of desired characters, but it is important to take care of this error case. So, after each scanf(), you should clear the input stream, which can be done using a method like:
void clear_input_stream()
{
int c;
while ((c = fgetc(stdin)) != '\n' && c != EOF);
}
You can call this method after each one of your scanf() calls.
You want to make sure that if you are writing into a char array you should not add the & in front of the variable you are writing to -- check this answer for an explanation (your compiler should warn you about that).
The last thing I want to mention is that you could also add a length to the %d specifier to limit reading an integer of a certain length. Your main() could look something like this:
int main()
{
printf("Enter your full name: ");
scanf(" %49[^\n]", data.name);
clear_input_stream();
printf("Enter your age: ");
scanf(" %3d", &data.age);
clear_input_stream();
printf("Enter your Phone Number (xxx) xxx-xxxx: ");
scanf(" %14[^\n]", data.phone);
clear_input_stream();
printf("Enter your street address: ");
scanf(" %49[^\n]", addr.street);
clear_input_stream();
printf("Enter your city: ");
scanf(" %24[^\n]", addr.city);
clear_input_stream();
printf("Enter your state abbreviation: ");
scanf(" %2[^\n]", addr.state);
clear_input_stream();
printf("Enter your zip code: ");
scanf(" %5d", &addr.zip);
clear_input_stream();
printf("Personal Data: \n %s\n %d\n %s\n %s\n %s\n %s\n %d\n", data.name, data.age, data.phone, addr.street, addr.city, addr.state, addr.zip);
return 0;
}
I am new to C language and trying to create a simple program to return name and age. I created a working function for returning the name but this does not work for returning the int.
The code I have now is:
int * GetAge(){
int Age;
printf("What is your age: ");
scanf(" %d", &Age);
int * returnedage = Age;
return returnedage;
}
This is GetName():
char * GetName(){
char Name[31];
printf("What is your name: ");
scanf("%s", Name);
char * returnedname = Name;
return returnedname;
}
The warning is on this line:
int * returnedage = Age;
It says:
incompatible integer to pointer conversion
initializing 'int *' with an expression of type 'int'; take
the address with &
I have tried:
int * returnedage * Age;
int * returnedage & Age;
//for strcpy I set the function as a char
char * returnedage;
strcpy(Age, returnedage);
None of these work.
I want to just get the name and age then in main I am printing the name and age with:
printf("Your name is %s and your age is %d", GetName(), *GetAge());
This does not have any errors.
So my expected Output is:
What is your name: Ethan
What is your age: 13
Your name is Ethan and your age is 13
What I actually get is:
What is your name: ethan
What is your age: 13
exit status -1
Please tell me if there is a basic solution for this.
Change your code to this:
int GetAge()
{
int Age;
printf("What is your age: ");
scanf(" %d", &Age);
return Age;
}
On main (Remove the * on the GetAge()):
printf("Your name is %s and your age is %d", GetName(), GetAge());
You have overcomplicated things. Read again the sources you are using to learn C to understand better what is going on.
Edited:
Change your GetName() to:
void GetName(char *name){
printf("What is your name: ");
scanf("%s", name);
}
Now on main:
char name[31];
GetName(name);
printf("Your name is %s and your age is %d", name, GetAge());
The reason for that is that C can not return an array of characters (this is what you are trying to accomplish somehow). Instead, you can give that function the memory address of a local variable which lives in main() and store the user's input into that variable.
Try:
int GetAge()
{
int Age;
printf("What is your age: ");
if (scanf("%d", &Age) != 1)
return -1; // return an error code if an integer couldn't be read
return Age;
}
Now call the function using GetAge().
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 :)
#include <stdio.h>
#include<stdlib.h>
int main()
{
char first[10], last[10], id[10];
int stdnum;
FILE *fptr;
fptr = fopen("C:\\c\\program.txt","w");
if(fptr == NULL)
{
printf("Error!");
exit(1);
}
else
{
//first name
printf("Enter name: ");
scanf("%s", first);
fprintf(fptr,"%s ",first);
//last name
printf("Enter last name: ");
scanf("%s ", last);
fprintf(fptr,"%s", last);
//id
printf("Enter id: ");
scanf("%s %d", id, &stdnum);
fprintf(fptr,"%s\n", id);
fprintf(fptr,"%d\n", stdnum);
fclose(fptr);
return 0;
}
}
I am writing a program that asks user for name, last name and student number. Student number format is lowercase letter followed by 8 digits.
when i run this code, I can enter first and last name. After i enter last name and hit enter in cygwin, the console gives me a blank line and I MUST type atleast a charcter or a number into it for it to display "enter student id".
Enter name: bob
Enter last name: jones
1
Enter id: a00998877
then the file output is something like this:
"bob jones11"
I want the output to be something like this:
"Bob Jones a00998877"
What am i doing wrong?
Why are you scanning in two inputs in this line:
scanf("%s %d", id, &stdnum);
id is a char array and will be able to hold alphanumeric input. You can remove the use of stdnum.
scanf("%s", id);
I have an example-with-stdin-stdout
char first[10], last[10], id;
id is a single char no need for an array.
scanf("%s", last);
Removed space after s
scanf(" %c%d", %id, &stdnum)
changed id to char from string
fprintf(fptr,"%c\n", id)
Same here, changed to char
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.