I'm having a bit of trouble passing multiple variables to scanf:
#include<stdio.h>
int main(void) {
char *name;
float weight;
printf("Please enter your name, then weight separated by a comma: ");
scanf("%s,%f", name, &weight);
printf("%s, your weight is %.2f!\n", name, weight);
return 0;
}
Please enter your name, then weight separated by a comma: Tommy,184
Tommy,184, your weight is 0.00!
What would be the proper way to do this, and why doesn't scanf detect the comma and pull the necessary values in their variables?
There are two mistakes in your code:
Wild pointer
char *name; is not a safe code. *name* is a wild pointer and it points to some arbitrary memory location and may cause a program to crash or behave badly. You should use an array like char name[10]. You can refer to this link.
Comma in scanf
You can check this thread for more details
The comma is not considered a whitespace character so the format specifier "%s" will consume the , and everything else on the line writing beyond the bounds of the array sem causing undefined behaviour
I've just tried to modify your code, you should consider removing the comma, and replace it with a space
printf("Please enter your name, then weight separated by a space: ");
scanf("%s %f", name, &weight); // your code should work properly
I think you should be using a char array for the name variable and if you want to store its adress you would assign it to a pointer, doesn't make sense the way you wrote it
float weight;
char[20] name;
scanf(" %s, %f", &name, &weight);
Can you give this a try?
int main()
{
int num;
char str[50];
scanf("%49[^,],%d", str, &num);
printf("%s %d",str, num);
return 0;
}
EDIT
Explanation:
We can read input up to a specific character/characters using %[^<Your character(s)>]. The number 49 here simply refers to the length of string you'd receive and the last character is a null. Adding specific length is up to you ;) This is just one example of getting input the way you asked.
Related
My issue is coming from the %c name input, I am getting an error that it is expecting type char * but has type char * [15] for the scanf function. I am also getting an error in the printf where the %c expects int but has type char *. I am still quite new at this so if it could be explained as simply as possible that would be amazing.
#include <stdio.h>
struct Student {
int StudentID;
char Name[15];
float Mark1;
float Mark2;
float Mark3;
} a;
int main() {
float ave;
printf("Please input Student's ID \n");
scanf("%d", &a.StudentID);
printf("Please input Student's name. \n");
scanf(" %c", &a.Name);
printf("Input Mark 1. \n");
scanf("%f", &a.Mark1);
printf("Input Mark 2. \n");
scanf("%f", &a.Mark2);
printf("Input Mark 3. \n");
scanf("%f", &a.Mark3);
ave = (a.Mark1 + a.Mark2 + a.Mark3) / 3;
printf("Student Detail\nStudent ID: %d\nName: %c\nMark 1: %.2f\n Mark 2: %.2f\n Mark 3: %.2f\nAverage: %.2f\n",
a.StudentID, a.Name, a.Mark1, a.Mark2, a.Mark3, ave);
return 0;
}
Your problem is related to the difference between an array of chars and a single char.
The %c format only reads in one character at a time.
If you wish to read a string of characters use %s and it will read until a whitespace. (Please make sure you don't try to read a name more than 14 characters long into your 15 character buffer)
In more depth, your char Name[15] is actually a pointer to a series of chars in memory. You are accidentally trying to change to pointer itself, instead of the chars that it points to. This is why the compiler expects a char * .
Instead if you truly meant to only read one char you could use
scanf(" %c", &a.Name[0]);
to place the character in the first block of the Name array.
If this is too complicated don't worry, it will all come eventually :)
For now I think %s will suffice.
You can use %14s to be extra safe.
Also don't forget to use %s in the final printf as well
In your scanf() call, %c tells scanf() to accept a single character. But your argument is a character array. C is a low-level language; it's not smart enough to realize you wanted a string (char array) as input. You have to tell it by using %s instead of %c.
Struggling on my homework assignment.
After I give inputs it stops rather than doing the random array and last print if anyone can help I would appreciate it
#include <stdio.h>
#include <stdlib.h>
int main() {
char name, color;
int age;
int *poer = &age;
char *p = &name;
char *ptr = &color;
printf("What is your name?\n");
scanf(" %s", &name);
printf("How old are you??\n");
scanf(" %d", &age);
printf("What is your favorite color?\n");
scanf(" %s", &color);
char *story[5] = {"old volkswagen beetle", "singlet", "quater", "left sock",
"blackberry bold ninek"};
srand(time(0));
printf("My pal right here %s is %d years old I feel like we have been coding"
" together for a hundred years now I always wonder where the time has gone"
" One thing I have wanted to know is why they love "
"their %s %s so much I guess I might never know\n",
name, age, color, story[rand() % 5]);
return 0;
}
Plenty of errors can be caught by enabling the compiler warnings.
Here is the list of all problems with their proper solutions:
For this line:
srand(time(0));
You need to include the time.h library, otherwise, you will get an implicit definition of time() error message. Passing NULL to time() is a good practice.
As people figured out:
char name, color;
A single char-type cannot hold more than a single ASCII character. Thus, you need to formulate a sequence of characters, called character array. In C99, variable-length arrays are supported. Still, using a macro constant for defining string lengths during compile-time makes the code good. Replace it:
#define MAX_LENGTH 128 // Prefer your length
...
char name[MAX_LENGTH], color[MAX_LENGTH];
In the similar lines:
printf("What is your name?\n");
scanf(" %s", &name);
scanf() function stops reading further input after whitespace. For example, if you input: John Doe, the name will only store John, Doe becomes truncated. Also, never put an ampersand sign for a char-array in scanf().
fgets() is considered safer than this, since it accepts a limited number of characters defined in its second argument. Replace the scanf():
// ... , sizeof name, ... --- is also applicable here
if (fgets(name, MAX_LENGTH, stdin) == NULL) {
// Do something when input's invalid
}
// Input is okay
Note: The fgets() leaves a trailing newline character as soon as it stops receiving user input. To discard this, use this after its usage:
name[strcspn(name, "\n")] = 0;
After applying all these changes, you will not get any further problems.
the following code will be fine
char name[256];
char color[256];
int age;
printf("What is your name?\n");
scanf(" %s", name);
printf("How old are you??\n");
scanf(" %d", &age);
printf("What is your favorite color?\n");
scanf(" %s", color);
in your code, a char was casted to char* the variable will hold data larger than it could. the the stack was ruined
You used char for string which is wrong. You need a char array to store the strings.
Try this:
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#define STR_MAX 300
int main()
{
char name[STR_MAX];
char color[STR_MAX];
int age;
printf("What is your name?\n");
scanf(" %s", name);
printf("How old are you??\n");
scanf(" %d", &age);
printf("What is your favorite color?\n");
scanf(" %s", color);
char *story[5] = {"old volkswagen beetle","singlet","quater","left sock","blackberry bold ninek"};
srand(time(0));
printf("My pal right here %s is %d years old I feel like we have been coding together for a hundred years now I always wonder where the time has gone One thing I have wanted to know is why they love their %s %s so much I guess I might never know\n",name, age, color, story[rand()%5]);
return 0;
}
your program terminated abnormally because stack was ruined.
char name; //has only 1 byte space
char color; //has only 1 byte space
but these two variables was used to hold data larger than they can. then data was written to some place else which will cause the stack overflow.
code like this will be ok
char name[256];
char color[256];
Very simple C printing question!
#include <stdio.h>
#include <conio.h>
int main() {
int Age = 0;
printf("Enter your Age\n");
scanf("%d",&Age);
char Name;
printf("Enter your Full name\n");
scanf("%s",&Name);
printf("My name is %s and I am aged %d" ,&Name,Age);
return 0;
}
When I input "blah" and 1, for some reason this returns:
"My name is Blah and I am aged 1929323232"
I presume I am misunderstanding a data format in either the scanf or the printf functions but can't work it out.
The problem is because of line
char Name;
Name is of type char. That means that it is supposed to store only one character. As a result
1. The scanf() is not able to store the input text properly (this will result in a crash in most cases or other undefined behaviour depending on the system - which judging by the output you provided is what you got)
2. (if the code didn't crash) Treating Name as a string with the %s argument in printf() essentially outputs garbage.
The type that corresponds to strings in C is char * (or char[]). Essentially, changing Name to some statically allocated char-array while performing the necessary changes in the next lines should fix your error:
char Name[256]; //allocated 256 bytes in Name array
printf("Enter your Full name\n");
scanf("%s",Name); // removed & before Name
printf("My name is %s and I am aged %d" ,Name,Age); // same here
You could also opt to go with a dynamically allocated string of type char * but I guess that's a different topic altogether.
As a general suggestion, I think you should look at pointers more closely. Especially in C, almost all string operations involve being aware of pointer mechanisms.
My goal with this program is to incorporate the users inputs into a sort of interactive/randomized story but I'm not sure how I'm supposed to get the inputs from the users to fit between *ptrDescription, *ptrBeginning, *ptrMiddle, and *ptrEnd. Any help would be much, much appreciated!
#include <stdio.h>
#include<stdlib.h>
#include<time.h>
#include <string.h>
#include <ctype.h>
int main(void){
int i;
char name[20];
char color[20];
int age;
char sentence[1];
//array of pointers to char arrays
char *ptrDescription[]={"the painfully handsome","the one and only","who seemed much older than"};
char *ptrBeginning[]={"was blissfully ignoring","could clearly see","had no idea"};
char *ptrMiddle[]={"the huge truck","the falling meteor","the bucket of milk","the mailman","the most powerful wizard"};
char *ptrEnd[]={"that was barreling toward them.","on the horizon."};
srand(time(NULL));
printf("Enter your first name: ");
scanf("%s", &name);
printf("\nEnter your age: ");
scanf("%d", &age);
printf("\nEnter your favorite color: ");
scanf("%s", &color);
for (i = 0; i < 1; i++)
{
//strcpy(sentence,ptrDescription[rand()%3]);
//strcat(sentence," ");
//strcat(sentence,ptrBeginning[rand()%3]);
//strcat(sentence," ");
//strcat(sentence,ptrMiddle[rand()%5]);
//strcat(sentence," ");
//strcat(sentence,ptrEnd[rand()%2]);
//strcat(sentence,".");
//sentence[0]=toupper(sentence[0]);
puts(sentence);
}
getch();
return 0;
}
EDIT:
I've edited a section of my code so that directly following for (i = 0; i < 1; i++) it now looks like this:
snprintf(sentence, sizeof sentence,"%s, %s %d year old, %s %s %s %s", name, ptrDescription[rand()%3], age,ptrBeginning[rand()%3], ptrMiddle[rand()%5], ptrEnd[rand()%2]);
There are tons of strange characters after the sentence in the output, like Japanese characters and stuff. I'm not sure why they're there, though. This is what it looks like exactly:
"Enter your first name: Justin
Enter your age: 20
Justin, the arrogant 20 year old, was purposefully ignoring the most powerful wizard that was barreling toward them. 汽$0HβHζ(ﾃﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌH・(DｷHH広$0陏&・汽$0ﾀHζ(ﾃﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌH WH・ H櫛H・t9HνHﾃ<"
Anyone know how I can get rid of them?
If you already have a name and an age, it's just a matter of inserting them into the correct place in sentence, right? So strcat(sentence, name) would work for name. age is a little trickier since you have to format the number first, and strcat won't do it for you. One solution would be to use sprintf(buf, "%d", age), and then concatenate buf (which is a scratch char array you would have to declare).
Any time you work with strings in C, you have to be concerned about having enough space in the target buffer. Your program can run out of space during both input and output. For the output, I would get rid of sentence altogether; since you just end up writing to stdout I would printf("%s", [part]) each part as you go along. For reading, scanf supports adding a length argument to the format string.
If you use one of the *printf functions, there are 2 things you must be careful about:
The arguments you pass are correct for the format string you use
Your buffer ends up null-terminated
Your current problem is with #1 - your format string promises 7 arguments to follow, but you only supply 6. snprintf grabs a "random" 7th value from the stack, interprets it as a char pointer, and copies whatever it finds there to sentence. You could see similar problems if your format string promised a char pointer but you placed an int in a given position. In this case the format string is a constant, so a smart compiler can validate that your format string matches the subsequent parameters. You'll want to get into the habit of taking compiler warnings seriously and not ignoring them.
The second point could be an issue if your sentence ended up bigger than your sentence buffer. If there is no room for a null-terminator, one won't be applied. You can check the return value of snprintf, or you can defensively always write a 0 to the last array position.
I'm trying to store a few values in a struct object and I want to repeat the prompt until the user types in "yes". I want to use a do-while loop for that. I'm already failing with the read-in of the first "last name". When I type in something, the program just stops (no error). I don't even use the do-while, since I'm not sure if it will work with my while() condition.
#include <ctype.h>
#include <stdio.h>
#include <string.h>
struct employeelist
{
char last[6];
char first[6];
int pnumber;
int salary;
};
int main()
{
struct employeelist employee[5];
char check;
//do
//{
printf("Hello. Please type in the last name, the first name, the personal number and the salary of your employees.\n");
printf("Last name: ");
scanf("%c", employee[1].last);
printf("First name: ");
scanf("%c", employee[1].first);
printf("Personal number: ");
scanf("%d", &employee[1].pnumber);
printf("Salary: ");
scanf("%d", &employee[1].salary);
printf("You have more employess (yes/no)?: ");
scanf("%c", &check);
//}while (scanf("yes"));
return 0;
}
Use %s as your format specifier if you're trying to get a string. You might also want to limit its length to 5, since that's how much space you have for last and first. So that would be %5s. Also, 5 characters is pretty short for a name.
Another comment: arrays in C are zero-based, so employee[1] is the second employeelist in your array. If you want to do this in a loop with an incrementing index, start at 0.
Hi when you read char array you must use scanf("%s", employee[1].last); %s but not %c
What do you think this code does?
scanf("%c", ....
%c indicates that scanf should only read ONE character.
One letter is not going to get you an entire name.
You need to switch to %s for starters.
First of all the first index to work
with will be '0',not 1.
wrong identifier for string,it
should be %s.
If you just want to iterate by
asking y/n then just change the
display message from yes/no to y/n
and also change that strange while
condition to check=='y'||check=='Y'.
The code will actually not work
after 5 iterations because you
initialized only 5 structures of
that type.Why don't you add that in
the loop?