I have written the following code to multiply two numbers and I see the result to be incorrect. I assume the data type is not holding the value right. But the answer its printing is incorrect.
#include<stdio.h>
main()
{
long int val1,val2;
val1=val2=1235;
char c = 'y';
switch(c)
{
case 'y' : printf("%20d",val1*val2);
break;
default: printf("invalid");
break;
}
return 0;
}
OUTPUT :
17897 //which is not the right answer
//it should actually be 1525225
Data type is holding the right value but you are using the wrong specifier. You should use %ld with long int.
printf("%20ld",val1*val2);
I double checked and it worked.
Related
Please give me some feedback on how to make my code better or more efficient. It should convert a decimal integer to binary.
#include <stdio.h>
binarydigits(int div, int dis)
{
int numit;
numit=0;
do
{
++numit;
div /= dis;
}
while (div!=1);
++numit;
return numit;
}
main()
{
int x, nb, i;
printf("\n Input an decimal integer number to be converted: ");
scanf("%d", &x);
fflush(stdin);
if (x==0 || x==1)
{
printf("\n\n %d in binary : %d", x, x);
}
else
{
printf("\n\n %d in binary : ", x);
nb = binarydigits(x, 2);
// the function 'binarydigits' returns how many binary digits are needed to represent 'x'
int remind[nb];
// an array of 'nb' elements is declared. Each element of this array will hold a binary digit
for(i=(nb-1) ; i>=0 ; --i, x/=2)
{
remind[i] = x%2;
}
//this 'for' structure saves the remainder of 'x/2' (x%2) in an element of the 'remind[nb]' array
for (i=nb ; i>0 ; --i)
{
printf("%d", remind[nb-i]);
}
//this 'for' structure prints the elements of the 'remind[nb]' array in increasing order
}
getch();
return 0;
}
Any tips on how to make this better would be nice.
Firstly, binarydigits should have a return type int. This is because you return an integer variable numit at the end of this function. Change your function header to:
int binarydigits(int div, int dis)
Secondly, the main() function needs to have a return type int by definition in C, and C++ for that matter. Without it, your compiler will produce a warning, something similar to:
main.c:18:1: warning: return type defaults to 'int' [-Wimplicit-int]
main()
^~~~
Here is a snippet from the the C11 standard (ISO/IEC 9899:2011) on the definition of the main() function:
The function called at program startup is named main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters: - Return Type of main()
int main(void) { /* ... */ }
Thirdly, you should remove fflush(stdin) because using the fflush() for stdint is undefined behavior as it is not a part of standard C. From C11 7.21.5.2, fflush works only with output/update stream, not input stream:
If stream points to an output stream or an update stream in which the most recent operation was not input, the fflush function causes any unwritten data for that stream to be delivered to the host environment to be written to the file; otherwise, the behavior is undefined. - fflush(stdin)
How to make my code better or more efficient?
My advice to you is to stop trying to learn C by trial-and-error method. You should obtain a good book and study it first. It is impossible to create a fast and efficient C program without mastering pointers, bitwise operators and memory manipulations.
Simply, to make your code fast, you should completelly delete your code (I am not going to list all of your bad-practice things here) and start understanding my example:
int main(void){
char *s = (char*)malloc(33);
char *t = s;
int a;
s += 33;
*s = 0;
printf("Enter a number: ");
scanf("%d", &a);
printf("That number in binary: ");
while(a){
*(--s) = a & 1 ? '1' : '0';
a >>= 1;
}
printf("%s\n", s);
free(t);
return 0;
}
Explanation: we have pointer (if you don't know pointers, well you should probably first learn them) s which points to the end of a string. While number from input (number a) is nonzero, we put its last binary digit in the string, decrease pointer and divide a integrally by 2 (this is a >>= 1 instruction). When a is 0, just print the string.
I have done my fair share of studying the C language and came across this inconsistency for which I cannot account. I have searched everywhere and reviewed all data type definition and relational syntax, but it is beyond me.
From the book C How to Program, there is a question to make a binary to decimal converter where the input must be 5-digits. I developed the follow code to take in a number and, through division and remainder operations, split it into individual digits and assign each to an element in of an array. The trouble arises when I try to verify that the number entered was indeed binary by checking each array element to see whether it is a 1 or 0.
Here is the code:
#include <stdio.h>
int power (int x, int y); //prototype
int main(void)
{
int temp, bin[5], test;
int n=4, num=0;
//get input
printf("%s","Enter a 5-digit binary number: ");
scanf("%d", &temp);
//initialize array
while(n>=0){
bin[n]=temp/power(10,n);
temp %= power(10,n);
n--; }
//verify binary input
for (test=4; test>=0; test--){
if ((bin[n]!=0)&&(bin[n]!=1)){
printf("Error. Number entered is not binary.\n");
return 0; }
//convert to decimal
while(n<=4){
num+=bin[n]*power(2,n);
n++; }
printf("\n%s%d\n","The decimal equivalent of the number you entered is ",num);
return 0;
}
//function definition
int power(int x, int y)
{
int n, temp=x;
if(y==0) return 1;
for(n=1; n<y; n++){
temp*=x; }
return temp;
}
Could someone explain to me why regardless of input (whether: 00000, or 12345), I always get the error message? Everything else seems to work fine.
Thank you for your help.
Update: If the if statement is moved to the while loop before. This should still work right?
Update2: Never mind, I noticed my mistake. Moving the if statement to the while repetition before does work given the solution supplied by sps and Kunal Tyagi.
After this
while(n>=0){
bin[n]=temp/power(10,n);
temp %= power(10,n);
n--; }
n is set as -1 so when you try to convert to decimal the statement bin[n] is actually bin[-1] so it returns you error.
One issue is that, while checking if the number is binary or not, you are returning at wrong place. You need to return only if the number is not binary. But you are returning outside the if condition. So your program returns no matter what the input is.
for (test=4; test>=0; test--){
if ((bin[test]!=0)&&(bin[test]!=1))
printf("Error, numbered entered was not binary.\n");
// Issue here, you are returning outside if
return 0; } //exit program
You can change that to:
for (test=4; test>=0; test--){
if ((bin[test]!=0)&&(bin[test]!=1)) {
printf("Error, numbered entered was not binary.\n");
// Return inside the if
return 0; // exit program
}
}
There is one more issue. Before you convert your number to decimal, you need to set n = 0;
//convert to decimal
n = 0; /* need to set n to zero, because by now
n will be -1.
And initially when n is -1, accessing
bin[-1] will result in undefined behavior
*/
while(n<=4){
num+=bin[n]*power(2,n);
n++; }
This looks like a homework, but your issue is in the brackets. More specifically line 23. That line is not part of the logical if statement despite the indentation (since that doesn't matter in C).
No matter what, the program will exit on test=4 after checking the condition.
Solution:
if ((bin[test]!=0)&&(bin[test]!=1)) { // << this brace
printf("Error, number entered was not binary.\n");
return 0; } } //exit program // notice 2 braces here
I am trying to make a C program for converting a given number in say base x, to base y. I chose to narrow it down upto base 20 (i.e. Base 2 to 20). When it comes to scanning a hexadecimal number (includes ABCDEF too, right?) for example, I am stuck. Please look at my program below:
/* NOTE: This program uses two step approach to convert a given number in any base (except base 10, in which case we will use only "toany()") to any other base*/
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int inum,ibase, obase;
int todec(); //function to convert to decimal from any base
int toany(int); //function to convert from decimal to any base
int exp(int,int); //used in other function
void main()
{
int num,choice;
char strr[100];
enum{A=10,B,C,D,E,F,G,H,I,J};
here:
printf("Enter the base (RADIX) of your number: ");
scanf("%d",&ibase);
printf("Enter the number in base %d: ",ibase);
scanf("%s",strr);
printf("Enter the base in which you want the output: ");
scanf("%d",&obase);
inum=atoi(strr);
switch(obase)
{
case 10:
num=todec();
printf("Output in base 10: %d\n",num);
break;
default:
if(ibase==10)
num=toany(inum);
else
num=toany(todec());
printf("Output in base %d: %d\n",obase,num);
break;
}
printf("WANNA DO IT AGAIN? If yes, Press 1 else press 0:");
scanf("%d",&choice);
if(choice==1)
goto here;
else
exit(0);
getch();
}
int exp(int p, int q)
{
int i,result=1;
for(i=1;i<=q;i++)
{
result=result*p;
}
return(result);
}
int todec()
{
int inumarr[100],dupnum=inum,i=0,counter,decnum=0;
while(dupnum!=0)
{
inumarr[i]=dupnum%10;
dupnum/=10;
i++;
}
for(counter=0;counter<i;counter++)
{
decnum=decnum+inumarr[counter]*exp(ibase, counter);
}
return(decnum);
}
int toany(int num)
{
int outnumarr[100],i=0,q,result=0;
while(num!=0)
{
outnumarr[i]=num%obase;
num=num/obase;
i++;
}
for(q=0;q<i;q++)
{
result=result+outnumarr[q]*exp(10,q);
}
return(result);
}
Thanks for reading! Now, I know it's definitely a mess where I tried to scan as a string and then applied atoi function on a string that might contain alphabets (like "19E" in base 16...which is 414 in base 10). So, I am looking for a decent solution which will allow the user of this program to enter any number like "19E" and my program will interpret that 'E' as 14 (AS DEFINED IN MY ENUM) and also a decent way to show an output of numbers like "19E" would be great.
Disclaimer: The code I've put into this answer is untested. I'm currently on a mobile device, so even compiling it is less convenient than usual. I will strive to include enough details for you to find your way past any (possible) errors, please point them out though... On another day I'll polish this post off by adding more checks (described at the end) and explain serialisation as well as deserialisation. As it stands, however, it seems you're just asking about deserialisation, so without further adeau:
Build a lookup table of some description containing each character from your base. For characters then you can (usually) get away with using string operations. For example:
unsigned char hex_digit[] = "00112233445566778899AaBbCcDdEeFf";
If you use strchr and some pointer arithmetic you can now find the offset of a character, divide by two to reduce it to a value within 0 .. 15, or modulo by two to discriminate between lowercase and uppercase.
You can devise any base like this, with a generic loop parsing the input to facilitate larger values...
size_t to_native_uimax(char *str, unsigned char *base, uintmax_t *value) {
size_t x, base_size = strlen(str);
uintmax_t v = 0;
for (x = 0; str[x]; x++) {
unsigned char *c = strchr(base, str[x]);
if (!c) break;
v *= base_size / 2;
v += (c - base) / 2;
}
*value = v;
return x;
}
Signage is a bit trickier to handle, but because we only need to handle the sign at the start of the string we can reuse the code above.
size_t to_native_imax(unsigned char *str, unsigned char *base, intmax_t *value) {
uintmax_t v = 0;
size_t x = to_native_uimax(str + !!strchr("-+", *str), base, &v);
*value = *str == '-' ? -(intmax_t)v : v;
return x;
}
Also note that this code isn't strictly portable; if it's possible that this might be deployed to a system that has negative zeros or signals on overflow more checks should precede the (intmax_t) conversion.
what is wrong with this im to make c code for total price with one of three ram options.et ramchoices=1or2or3 /this is where I know my problem lies but don't know how to fix it. the compiler needs a one to one relationship. how do I break this up and still show that ramchoice depends on what I pick and those choices affect price./
int main(void) {
/*declare variables*/
float baseprice;
float total;
float ramchoice;
baseprice=1029.48;
/*ramchoice can be 1 or 2 or 3*/
ramchoice=1||ramchoice=2||ramchoice=3;
total=baseprice+ramchoice;
/*initiate variable*/
scanf("%f",&ramchoice);
if("%f" (ramchoice==1))
{
total=baseprice+179.99;
}
else if("%f" ramchoice==2)
{
total=baseprice+94.99;
}
else if("%f" ramchoice==3)
{
total=baseprice+69.99;
}
printf("total is %f",total);
return 0;
}
There are several lines that are unlike any C code I've ever seen before.
Are you completely making up your own syntax and meaning?
Have you read any instructions/tutorials on C?
ramchoice=1||ramchoice=2||ramchoice=3;
Whatever you think this line does, it sets variable ramchoice to have value 1, then it stops.
if("%f" (ramchoice==1))
if does NOT take a string like "%f". if expects a boolean expression.
You're close to a boolean with ramchoice==1, but the "%f" at the start is so very wrong.
Seriously, you cannot make up rules to a language and pray that it works.
Here is my re-write that pretty much fixes your issues:
int main(void)
{
/*declare variables*/
float baseprice = 1029.48;
float total;
int ramchoice;
printf("Please enter a ram selection: 1, 2, or 3\n");
scanf("%d",&ramchoice);
switch(ramchoice)
{
case 1: total=baseprice+179.99; break;
case 2: total=baseprice+94.99; break;
case 3: total=baseprice+69.99; break;
default: printf("That was not a valid choice\n");
}
printf("total is %f",total);
return 0;
}
There's somethings in your code that make no sence (are not correct in C):
ramchoice=1||ramchoice=2||ramchoice=3; total=baseprice+ramchoice;
"%f" (ramchoice==1)
The first I don't understand why you are using it.
The second one gives me the idea that your not understanding how scanf works.
Also, you are declaring ramchoice as a float, but you are comparing it to another value (1, 2 or 3). Due to the representation of float values, this may not work. I mean, if you have float x = 1; and then do x == 1, it may return false. To fix this you can just change it to an int (in this case, in others you can use an error margin).
Here's a fix of your code:
int main(void) {
/*declare variables*/
float baseprice;
float total;
int ramchoice;
baseprice=1029.48;
/*initiate variable*/
scanf("%i",&ramchoice);
if(ramchoice==1)
{
total=baseprice+179.99;
}
else if(ramchoice==2)
{
total=baseprice+94.99;
}
else if(ramchoice==3)
{
total=baseprice+69.99;
}
printf("total is %f",total);
return 0;
}
Get rid of the "%f" in the if statements
Comment those 2 lines:
ramchoice=1||ramchoice=2||ramchoice=3;
total=baseprice+ramchoice;
and code should compile.
I am trying to write a program which calculates and prints the GC content of a string of DNA(which is input through a txt file). That is, the percentage of G's and C's in a string of DNA. Here is my function for the GC percentage:
void updateGCCount(char s[], int *gc, int *at) {
char c[MAXLENGTH];
int i,GCcount,ATcount;
float len,GCpercentage;
GCcount=0;
ATcount=0;
for(i=0;c[i]!='\0';++i)
{
if(c[i]=='G' || c[i]=='C')
{
++GCcount;
*gc=GCcount;
}
if(c[i]=='A' || c[i]=='T')
{
++ATcount;
*at=ATcount;
}
}
strcpy(c,s);
len=strlen(c);
GCpercentage=*gc/len;
printf("GC-content: %.2f\n",GCpercentage);
}
This is my function definition, and the part which is supposed to correctly print the GC percentage is what I am not sure about. Below is my main program which utilizes the input text file.
#include "genomics.h"
int main(){
char s[MAXLENGTH];
int gc, at;
scanf("%s",s);
printf("Sequence : %s\n",s);
updateGCCount(s, &gc, &at);
return 0;
}
Any help or advice on why I am not getting a correct value for the GCpercentage would be great. Thank you in advance
You're doing your tests on char array "c":
char c[MAXLENGTH];
...
for(i=0;c[i]!='\0';++i)
{
if(c[i]=='G' || c[i]=='C')
{
++GCcount;
*gc=GCcount;
}
if(c[i]=='A' || c[i]=='T')
{
++ATcount;
*at=ATcount;
}
}
If should be on s, the array that you passed in. The c array is probably superflous, you should be able to get the length from s as well
c is not initialize, so *gc and *at are not updated at all and they contain garbage..
here you should use s instead of c
for(i=0;c[i]!='\0';++i)
{
if(c[i]=='G' || c[i]=='C')
{
++GCcount;
*gc=GCcount;
}
if(c[i]=='A' || c[i]=='T')
{
++ATcount;
*at=ATcount;
}
}
That's a strongly un-idiomatic program. Consider the following.
#include <stdio.h>
#include <stdlib.h> /* for exit(3) */
float count_gc(const char* s)
{
You have no need to pass information back via variables passed in by reference. Functions return values -- typically 'the answer'.
You're simply scanning the content of the argument string s, so there's no need to copy it anywhere.
As others have pointed out, you were scanning the contents of the array c[] before you copied anything in to it -- you were counting 'G' and 'C' in a (probably large) random block of memory. Keeping things simple avoids mistakes like that.
int nvalid = 0;
int gccount = 0;
float result;
for (; *s != '\0'; s++) {
Although the for loop you wrote isn't wrong, it's somewhat un-idiomatic. Here, we examine the character pointed to by the pointer s, and then increment the pointer, until we find ourselves pointing at the \0 that terminates the string. Yes, this means we 'lose' the initial value of the argument, but we don't need it after the loop, so that doesn't matter.
switch (*s) {
A switch is a more natural construction here. You're looking for a small set of possible values that *s (that is, the character the pointer is currently pointing at) may have.
case 'G':
case 'C':
nvalid++;
gccount++;
break;
case 'A':
case 'T':
nvalid++;
break;
default:
/* unexpected character -- ignore it */
break;
Every switch statement should have a default clause -- one should always think of what's supposed to happen if none of the case clauses match. In this case, we just ignore this character.
}
}
if (nvalid == 0) {
fprintf(stderr, "No valid letters found!\n");
result = 0.0;
} else {
/* Multiply by 1.0 to convert integer gccount to a float */
result = 1.0*gccount / nvalid;
}
return result;
We return the result to the caller rather than printing it out inside the function. Functions shouldn't 'chatter', but leave all of the I/O in one place, typically leaving the main function (or something higher up) to look after that.
}
int main(int argc, char** argv)
{
if (argc != 2) {
/* Give the user a hint on how to call the program */
fprintf(stderr, "Usage: gcat <string>\n");
exit(1);
}
printf("Sequence GC-content = %g\n", count_gc(argv[1]));
}
I run that with:
% cc -o gcat gcat.c
% ./gcat "GCAT ATx foo"
Sequence GC-content = 0.333333
%
With C, it's very easy to tie yourself in knots, very quickly. Aim for simplicity always.