I have no idea why this code is not working. Could someone please help me? (I want a simple version-fix if it's possible, because I've only started to study C a couple of weeks ago.)
#include <stdio.h>
#include <string.h>
int main ()
{
char *name="alina";
char *input;
printf ("what's your name? \n");
scanf ("%s",&input);
if (input=="alina")
printf("your name is %s good job!\n ",&name);
if (input!="alina")
printf("are you sure? open the program again and insert the correct name");
while (1);
}
you need to apply following changes:
char input[256];
scanf("%s", input);
printf("your name is %s good job!\n ", name);
to compare strings use strcmp - have a look at documentation of this function.
be careful with scanf - think about how long string you can store there and what can go wrong. Have a look at scanf_s
You did some errors. First, if you want to insert a string, you can use the %s, but you have to use an array of char in which you can store that string. That's, you have to write something like this.
char input[100];
scanf("%s", input);
and then it'll work. This snippet of code means: I need to insert (store) a string. So first I create a place in which I can store it (pay attention that the string must be maximum of 99 characters; my array has size 100, but the last character is used to represent the end of the string), and then I use the scanf to write what I want.
The second error is that if you want to compare two strings, you can't simply use the == or !=, like when you do with numbers. The string.h library lets you use the strcmp function, like this:
if (strcmp(name, input) == 0) // this means the strings are equal
...
else
...
remembering that
char* name = "Alina";
char input[100];
Eventually, here you're an inspected version of your code:
#include <stdio.h>
#include <string.h>
int main() {
char* name = "Alina";
char input[100];
printf("What's your name?\n");
scanf("%s", input);
if (strcmp(name, input) == 0)
printf("Your name is %s good job!\n", name);
else
printf("Are you sure? Open the program again and insert the correct name\n");
return 0;
}
The while(1) at the end of your code is absolutely dangerous, because it starts an infinite loop that never ends, crashing your program. You want to definitely remove it!
Related
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];
The problem is about taking input of strings x number of times using an array of pointers. x is the value entered by the user. I wrote the following code for the same. But the program is taking only x-1 inputs.
I have inserted fflush(stdin) because I think the scanf is consuming an enter first but I don't know from where.
I have tried using gets but with no use.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
//code to take input in an array of pointers
int x,i,j,length;
char ch[50],*t;
printf("How many names you want to sort:\n");
scanf("%d",&x);
char *names[x];
char *p;
printf("Enter the names:\n");
for(i=0;i<x;i++)
{
fflush(stdin);
scanf("%[^\n]s",ch);
length = strlen(ch);
p = (char *)malloc((length+1) * sizeof(char));
strcpy(p,ch);
names[i] = p;
}
return 0;
}
Why bother with complex format strings if you don't have to? Use fgets.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void err(const char * msg) {
fprintf(stderr, msg);
exit(1);
}
int main()
{
int x,i;
char ch[50];
printf("How many names you want to sort:\n");
if(!fgets(ch, 50, stdin)) err("Error reading line");
if(sscanf(ch, "%d",&x) != 1) err("Could not read integer");
// Better than using VLA
char **names = malloc(x*sizeof(*names));
if(!names) err("Error allocating names");
printf("Enter the names:\n");
for(i=0;i<x;i++) {
if(!fgets(ch, 50, stdin)) err("Error reading line");
ch[strcspn(ch, "\n")] = 0; // Remove newline
if(!(names[i] = strdup(ch))) err("Error duplicating string");
}
for(int i=0; i<x; i++)
printf("name %d: %s\n", i, names[i]);
}
Whenever a function has a return value that may indicate an error you should ALWAYS check it, and here that is the case for malloc, fgets, strdup and sscanf and. Read the documentation to find out what it actually returns to see how to check for errors. sscanf returns the number of successful assignments, and the other three returns a pointer which is NULL on failure.
You wrote in the comments that you are learning from the book "Let us C". A better fitting title would be "How to not code C". I've had a quick look at it and, it is really really bad. Apart from teaching very outdated C, it also teaches very bad habits in general, and many of the things you can read is completely WRONG. Actually, the majority of questions about C can be traced to that book, or at least could have. Two prime examples is that it consistently avoids very important stuff, such as error checking functions like scanf and malloc. I have not read every line, but I think it does not even mention how to error check scanf even once. It also uses the function gets which is not only deprecated but completely removed from newer C standards because it is so dangerous. It also says that you can modify a string literal, which is undefined behavior in C.
I'm running a while loop so the user can constantly enter expressions, until they indicate they want to quit the program. I'm using strcmp() to compare two strings so as soon as they enter quit the program will stop. But the program keeps going, any Ideas?
#include <stdio.h>
#include <string.h>
int main()
{
int min12=0;
char opper;
int x=0;
int min13;
char *Repeatprog="cont";
char *Repeatprog1="quit";
while (strcmp(Repeatprog,Repeatprog1))
{
printf("enter the integer number \n");
scanf( "%d %c %d", &min12, &opper, &min13);
printf("%d %c %d\n", min12, opper, min13);
printf("Type the word quit to end program\n");
scanf("%s", Repeatprog);
}
printf("Good Bye");
return 0;
}
Remember always that an Array is a Pointer to the first object of the array.
And secondly, in your call to scanf() you only read a character. Not a whole string (represented by %s in C)
So in conclusion, your call to scanf() shouldn't have a pointer and should have a string instead of a character.
scanf("%s", Repeatprog);
or simply
gets (Repeatprog);
EDIT :
As the commenter #EOF said, gets() is not a good idea since it can lead to Undefined Behaviour. That's because the program can read more characters than it should have and lead to overflow, thus it isn't secure.
So I recommend using char *fgets(char *str, int n, FILE *stream)
Note:
Also, your code is using string literals. So if you make any attempt to change the content of the char pointer then it will lead to Undefined Behaviour.
For this note, please thank the guys below me [comments]. I made a huge mistake and I'm sorry.
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 ask the user to type in a string so I will print the length of the string. My code is built succeeded. However, when I entered a word and pressed 'enter', the program keeps running. I had to enter a second word, then the length of the first string displays. I'm confused at argv[1]. Can someone give me some tips and hint on how to fix this? Thanks in advance for your time.
Please note that I'm not allowed to use any string function.
int main(int argc, char* argv[]){
char* s=argv[1];
char input[256];
s = input;
printf("Please enter a string: ");
scanf("%s\n", s);
int str_length = 0;
while (s[str_length] != '\0')
{
str_length++;
if (s[str_length] == '\0') break;
}
printf("%d\n", str_length);
return 0;
}
argv[] is the array that holds your command line parameters, argv[1] is the first one (other than the command representation itself in argv[0], of course). You assign it to s then immediately overwrite s, so it's not really needed here.
The reason you're having to enter two lines is the \n at the end of your input format. It requires whatever can match the format string followed by a newline, hence the %s\n is eating your first newline so that scanf has to go back for another one.
%s on it's own will fix that problem but introduce another one if what you're after is a full line of input - it will only read up to the first whitespace. For a proper line input function, see here.
It does full line input with protection against buffer overflows and detection/cleanup of lines that were too long, something sorely missing in the scanf("%s") input method.
In addition, the if statement within the while is superfluous here since the while itself will catch the end of the string, and it makes little sense to have both input and s refer to the same array (it would make sense if you changed s at some point, but that's not happening here).
So, a variant without the advanced line iput function could be as simple as:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main (int argc, char* argv[]) {
char input[256];
int str_length = 0;
printf ("Please enter a string: ");
scanf ("%s", input);
while (input[str_length] != '\0') /* Or consider using strlen() */
str_length++;
printf ("%d\n", str_length);
return 0;
}
If you enter Hello, you'll see that it prints out 5 immediately. You'll also see 5 if you enter Hello Pax, which is one reason to choose the advanced input function (the other, very important, reason is to avoid introducing a buffer overflow vulnerability into your program).
argv is used to pass process arguments. this means you can run your program like
a.out 123 abcd
argv[1] will already be assigned to a value of 123. you do not have to read it as you are doing now. in your current code , you are overwriting the reference of s to a new string after assigning it with argv[1].
This is what you want
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define MAX 100
int main(int argc, char* argv[]){
char tmp[MAX];
printf("Plese enter a string:\n");
scanf("%s", tmp);
printf("the length is %d\n", strlen(tmp));
return 0;
}
The argv and argc is another thing like this:
argv[0] = a.exe
argv[1] = 12345
int main(int argc, char* argv[]){
printf("the length is %d\n", strlen(argv[1]));
return 0;
}
/*-------------------- Output --------------------
> a.out 12345
the length is 5
------------------------------------------------*/
Just change the scanf line:
scanf("%s", s);
As you have it (scanf("%s\n", s);), it requires a sequnce of characters ending in \n, then another \n to indicate when the input is finished.