I'm having difficult while trying to install a cross compiler under linux.
I downloaded a toolchain that seems to be ok but what's next ? what orders I need to type in the console to make it to be installed ?
The purpose is to convert a C code to MIPS (little endian) code.
Actually I need it once for only 2 codes, so if someone can only show me those codes in MIPS, I will be more than happy...
first code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIZE 128
int main ()
{
char mychar , string [SIZE];
int i;
int count =0 ;
printf ("Please enter your string: \n\n");
fgets (string, SIZE, stdin);
printf ("Please enter char to find: ");
mychar = getchar();
for (i=0 ; string[i] != '\0' ; i++ )
if ( string[i] == mychar )
count++;
printf ("The char %c appears %d times\n" ,mychar ,count);
return 0;
}
second code:
#include <stdio.h>
#include <string.h>
void SIFT(int x_arr[ ], int y_arr[]);
int main ()
{
int x[20] = {0} , y[20] = {0};
int m=0,temp=0,curr=0,i=0,j=0;
printf("Please enter your numbers now:\n\n");
/*enter numbers one by one. if x[i+1] value < x[i] value, err msg.
when user want to end the series he must enter '0' which means end of string (it wont included in x[]) */
while ( (scanf("%d",&temp) ) != 5 )
{
if (temp >= curr)
{
x[i] = temp;
curr = temp;
i++;
}
else
{
printf("The numbers are not at the right order !\n\nProgram will now terminate...\n\n");
}
}
SIFT(x,y);
for (i=0 ; y[i]=='0' ; i++) /*strlen(y) without ('0')'s includes*/
m++;
/*Prints m , y's organs*/
printf("\n\nm = %d",m);
printf("Y = ");
while (y[j]!='0')
{
printf ("%d ,",y[j]);
j++;
}
return 0;
}
void SIFT(int x_arr[ ], int y_arr[])
{
int i=0,j=0;
while (x_arr[i] != '0')
{
if (x_arr[i] == x_arr[i+1]) /*if current val. equals next val. -> jump dbl at x_arr*/
{
y_arr[j] = x_arr[i];
i+=2;
j++;
}
else
{
y_arr[j]=x_arr[i];
i++;
j++;
}
}
}
You may want to try using crosstool-NG to build and install a toolchain for your system.
I'm not sure if this will help you but I've compiled your source files using ecc (http://ellcc.org) and got:
http://pastebin.com/keDPEcsc
and
http://pastebin.com/zQBsMVfS
Hope that helps.
you need to build you own version with toolchain - it is not a compiler on it's own, it will make one. look at http://www.kegel.com/crosstool/crosstool-0.43/doc/crosstool-howto.html
Related
I wrote two sample programs to check for a palindrome string. But in both I am getting output like, its not a palindrome number. What I am missing?
I strictly assume somehow code is executing my if statement and put flag in to 1. May be because of that length calculation. Anyone has a better idea?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <conio.h>
int main(void) {
setbuf(stdout,NULL);
char name[100];
int i,length,flag=0,k;
printf("Enter your name");
/*scanf("%s",name);*/
gets(name);
length=strlen(name);
for(i=0;i<=length-1;i++)
{
for(k=length-1;k>=0;k--)
{
if(name[i]!=name[k])
{
flag=1;
break;
}
}
}
if(flag==0)
{
printf("Give word is a palindrome");
}
if(flag==1)
{
printf("This is NOT a palindrome word");
}
return 0;
}
and
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <conio.h>
int main(void) {
setbuf(stdout,NULL);
char name[100];
int i,length,flag=0;
printf("Enter your name");
/*scanf("%s",name);*/
gets(name);
length=strlen(name);
for(i=0;i<=length/2;i++)
{
if(name[i]!=name[length-1])
{
flag=1;
}
}
if(flag==0)
{
printf("Give word is a palindrome");
}
if(flag==1)
{
printf("This is NOT a palindrome word");
}
return 0;
}
First Algorithm
The algorithm you are using in the first program involves comparing each letter to every other letter which does not help in determining if the number is a palindrome and it does not seem fixable.
Second Algorithm
The problem with the second approach, however, is you are always comparing name[i] to name[length]. Instead change it to length-i-1. This will start comparing from length-1 and decrement the length of the character by 1 for every next iteration:
for(i = 0;i <= length / 2;i++)
{
if(name[i] != name[length-i-1])
{
flag=1;
break;
}
}
gets() and buffer overflow
Do not use gets. This method is susceptible to a buffer overflow. If you enter a string longer than 100 characters, it will result in undefined behavior. Use fgets instead for deterministic behavior:
fgets(name, sizeof(name), stdin);
This takes in the size of the buffer and only reads up to sizeof(name) characters.
Full code
Ideally, you should consider wrapping the logic to check if the string is a palindrome in a function:
int is_palindrome(char*);
int main(void)
{
char name[100];
setbuf(stdout,NULL);
printf("Enter your name");
fgets(name, sizeof(name), stdin);
if(is_palindrome(name))
{
printf("The given word is a palindrome");
}
else
{
printf("This is NOT a palindrome word");
}
return 0;
}
int is_palindrome(char* name)
{
int length = strlen(name);
int flag = 0, i;
for(i = 0;i <= length / 2; i++)
{
if(name[i]!=name[length-i-1])
{
return 0;
}
}
return 1;
}
There is plenty wrong with both your attempts. I strongly suggest using a debugger to investigate how your code works (or doesn't).
Your first attempt performs length2 (incorrect) comparisons, when clearly only length / 2 comparisons are required. The second performs length / 2 comparisons but the comparison is incorrect:
name[i] != name[length-1] ;
should be:
name[i] != name[length - i - 1] ;
Finally you iterate exhaustively when you could terminate the comparison as soon as you know they are not palindromic (on first mismatch).
There may be other errors - to be honest I did not look further than the obvious, because there is a better solution.
Suggest:
#include <stdbool.h>
#include <string.h>
bool isPalindrome( const char* str )
{
bool is_palindrome = true ;
size_t rev = strlen( str ) - 1 ;
size_t fwd = 0 ;
while( is_palindrome && fwd < rev )
{
is_palindrome = (str[fwd] == str[rev]) ;
fwd++ ;
rev-- ;
}
return is_palindrome ;
}
In use:
int main()
{
const char* test[] = { "xyyx", "xyayx", "xyxy", "xyaxy" } ;
for( size_t t = 0; t < sizeof(test)/sizeof(*test); t++ )
{
printf("%s : %s palindrome\n", test[t],
isPalindrome( test[t] ) ? "Is" : "Is not" ) ;
}
return 0;
}
Output:
xyyx : Is palindrome
xyayx : Is palindrome
xyxy : Is not palindrome
xyaxy : Is not palindrome
Try this
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
char text[100];
int begin, middle, end, length = 0;
printf("enter the name: ");
scanf("%s",text);
while ( text[length] != '\0' ){
length++;}
end = length - 1;
middle = length/2;
for ( begin = 0 ; begin < middle ; begin++ ) {
if ( text[begin] != text[end] ) {
printf("Not a palindrome.\n");
break;
}
end--;
}
if( begin == middle )
printf("Palindrome.\n");
return 0;
}
The problem with the first piece of code is you are comparing it more than required, compare it with length-i-1.
The main problem with the second code is you are comparing it with only the last letter of a word.
Hope you understood your mistake
I am trying to take a user inputted string and look at each code to see if it appears in another string of strings. So far my code works.
If the word is successfully found then the alpha representation is to be added to an array that will eventually be printed, but only if all codes were found.
I am having issues with what gets stored in my array that is going to be printed.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef char *string;
typedef char *alpha;
int main(void)
{
string morse[4]={".-", "-...","----.", ".."};
string alpha[4]={"A", "B", "9", "I"};
char prntArr[50];
char *input;
char *hold;
input = malloc(200);
hold = malloc(50);
int i=0;
int j=0;
int ret;
int x;
int w=0;
int z=0;
printf("please enter a string\n");
scanf("%[^\n]",input);
do{
if (input[i] !=' ')
{
hold[j] = input[i];
j++;
}
else
{
hold[j]='\0';
for (x=0;x<4;x++)
{
printf("value of x %d\n",x);
ret = strcmp(morse[x], hold);
if (ret==0)
{
printf("%s\n",alpha[x]);
prntArr[w]=*hold;
w++;
x=4;
}
else
{
ret=1;
printf("invalid Morse code!");
}
}
j = 0;
}
i++;
}while(input[i] !='\0');
for (z=0;z<50;z++)
{
printf("%c",prntArr[z]);
}
return 0;
free(input);
}
The problem you asked about is caused by the way prntArr is used in the program. It really should be an array of character pointers into the alpha array. Instead, it's manipulated as an array of characters into which the first character of each morse code element is stored. And when it's printed, the variable that tracks how much of the array is used is simply ignored.
Another problem is that your code uses spaces to break the codes but there won't necessarily be a space at the end of the line so a code might get missed. In the program below, I switched out scanf() for fgets() which leaves a newline character on the end of the input which we can use, like space, to indicate the end of a code.
Other problems: you print the invalid Morse code message at the wrong point in the code and you print it to stdout instead of stderr; you remember to free input but forget to free hold; you put code after return that never gets called.
Below is a rework of your code that addresses the above problems along with some style issues:
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main(void)
{
char *morse[] = {".-", "-...", "----.", ".."};
char *alpha[] = {"A" , "B" , "9" , "I" };
char *print_array[50];
int print_array_index = 0;
char hold[50];
int hold_index = 0;
char input[200];
int i = 0;
printf("please enter a string: ");
fgets(input, sizeof(input), stdin);
while (input[i] !='\0') {
if (input[i] ==' ' || input[i] == '\n')
{
hold[hold_index] = '\0';
bool found = false;
for (int x = 0; x < sizeof(morse) / sizeof(char *); x++)
{
if (strcmp(morse[x], hold) == 0)
{
print_array[print_array_index++] = alpha[x];
found = true;
break;
}
}
if (!found)
{
fprintf(stderr, "invalid Morse code: %s\n", hold);
}
hold_index = 0;
}
else
{
hold[hold_index++] = input[i];
}
i++;
}
for (int x = 0; x < print_array_index; x++)
{
printf("%s ", print_array[x]);
}
printf("\n");
return 0;
}
SAMPLE RUNS
> ./a.out
please enter a string: ----. -... .- ..
9 B A I
>
> ./a.out
please enter a string: .- --- ..
invalid Morse code: ---
A I
>
This is the problem I'm trying to solve:
Input:
First line contains N, the size of the string.
Second line contains the letters (only lowercase).
Output:
Print YES if all vowels are found in the string, NO otherwise.
Constraints:
The size of the string will not be greater than 10,000. 1 ≤ N ≤ 10000
The following code I wrote is always showing NO.
#include <stdio.h>
#include<conio.h>
int main()
{
int a,b,c=0,d=0,e=0,f=0,g=0,i;
char string[10000];
scanf("%d",&a);
scanf("%s",string);
for(i=0;i<a;a++)
{
if(string[i]==('a'))
c=1;
if(string[i]==('e'))
d=1;
if(string[i]==('i'))
e=1;
if(string[i]==('o'))
f=1;
if(string[i]==('u'))
g=1;
}
if((c==1)&&(d==1)&&(e==1)&&(f==1)&&(g==1))
printf("YES");
else
printf("NO");
return 0;
getch ();
}
Here is an infinite loop that causes a problem:
for(i=0;i<a;a++)
You should increment i, instead of a (length of a string). If you fix this one char in loop statement, the program will run well at all. Anyway, I changed your code a bit to be more readable. Take a look if you want, just for your information, sir:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int len, a=0, e=0, i=0, o=0, u=0, it;
char string[10000];
scanf("%d", &len);
scanf("%s", string);
for(it=0;it<len;it++)
{
if(string[it]=='a') a = 1;
else if(string[it]=='e') e = 1;
else if(string[it]=='i') i = 1;
else if(string[it]=='o') o = 1;
else if(string[it]=='u') u = 1;
}
if(a && e && i && o && u) printf("YES\n");
else printf("NO\n");
system("PAUSE");
return 0;
}
I assume you are running your program under Windows, so instead of conio's getch() try to use system("PAUSE") or just even better way to do this (for both Windows for UNIX): getchar()
I've renamed all of your variables, but otherwise left the code the same.
#include <stdio.h>
#include<conio.h>
int main()
{
int foundA = 0, foundE = 0, foundI = 0, foundO = 0, foundU = 0;
int i, length;
char string[10000];
scanf("%d", &length);
scanf("%s", string);
for(i=0; i<length; length++)
{
if(string[i]==('a'))
foundA=1;
else if(string[i]==('e'))
foundE=1;
else if(string[i]==('i'))
foundI=1;
else if(string[i]==('o'))
foundO=1;
else if(string[i]==('u'))
foundU=1;
}
if((foundA==1)&&(foundE==1)&&(foundI==1)&&(foundO==1)&&(foundU==1))
printf("YES");
else
printf("NO");
return 0;
getch ();
}
Looking the the for-loop condition for(i=0; i<length; length++), I think it's pretty clear what's wrong. Instead of incrementing the counter, you're incrementing the length of the string. Eventually, the counter overflows to a negative number, and the loop terminates without ever looking at a character besides the first one. The lesson here is to name your variables properly.
If you want to be picky, then signed integer overflow is undefined behavior, but for most systems, INT_MAX + 1 will be INT_MIN.
This program can be done in more simpler way other as below.
#include <stdio.h>
#include<conio.h>
int main()
{
int i, flag = 0;
char string[10000], *ptr;
char cmp[] = "aeiou";
printf("Please enter string = " );
scanf("%s", string);
i = 0;
while(cmp[i])
{
ptr = string;
while(*ptr)
{
if(cmp[i] == *ptr)
break;
ptr++;
}
if(*ptr != cmp[i++])
{
flag = 1;
break;
}
}
if(flag == 1)
printf("NO");
else
printf("YES");
}
In this program I have used just one flag instead of 5 flags. Always try to write simple code rather then using unnecessary variable and flags.
I wanted to make a C program that finds the numbers in the input array and then multiplies it all together, I made the program but it got an issue that I don't know, can anybody help me with it!?
Here is the code!
#include <stdio.h>
#include <stdlib.h>
int main ()
{
char t[10];
int n, z;
n = 0;
printf ("please enter a code: \n");
scanf ("%s", t);
while (n != '\0')
{
if (isdigit (t[n] == 0))
{
n++;
}
else
{
z = t[n];
z *= z;
}
}
printf ("%d", z);
}
Here is updated code. There is a comment for each bug that needed correction.
(Note that the comment describes the intention of the corrected code, it doesn't describe the bug.)
int temp;
z=1; // Initialize z
printf ("please enter a code: \n");
scanf ("%s", n);
while (t[n] != '\0') { // While there are more characters in the string
if (isdigit (t[n])) { // Check if the character is a digit
temp = t[n] - '0'; // Convert character digit to corresponding number.
z *= temp;
}
n++;
}
Your first problem is that you don't actually use t in your while loop. Your while loop only uses n which is set to 0 and never modified.
Your second problem is that you may be better off to use scanf("%d", &number); to scan numbers straight away.
z should be initialized to 1. and remove "z = t[n];"
#include <stdio.h>
#include <string.h>
main()
{
char a[5] ;
int b=1, n=0,m=0;
scanf("%s",a);
while (n <5 )
{
if (!isdigit(a[n]))
{
n++;
m++;
}
else{
b *= (a[n]-'0');
n++;
}
}
if(m==5) b=0;
printf("%d\n",b);
}
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Program not doing what it should - C
Hello,
Following program only read numbers from input and do stop when rull is violated, but 1 big problem is that it doesn't stop reading numbers and worse, don't print on screen what it should.
The code:
#include <stdio.h>
#include <string.h>
void SIFT(int x_arr[ ], int y_arr[]);
int main ()
{
int x[20] = {0} , y[20] = {0};
int m=0,temp=0,curr=0,i=0,j=0;
printf("Please enter your numbers now:\n\n");
/*enter numbers one by one. if x[i+1] value < x[i] value, err msg.
when user want to end the series he must enter '0' which means end of string (it wont included in x[]) */
while ( (scanf("%d",&temp) ) != '0' )
{
if (temp >= curr)
{
x[i] = temp;
curr = temp;
i++;
}
else
{
printf("The numbers are not at the right order !\n\nProgram will now terminate...\n\n");
}
}
SIFT(x,y);
for (i=0 ; y[i]=='0' ; i++) /*strlen(y) without ('0')'s includes*/
m++;
/*Prints m , y's organs*/
printf("\n\nm = %d",m);
printf("Y = ");
while (y[j]!='0')
{
printf ("%d ,",y[j]);
j++;
}
return 0;
}
void SIFT(int x_arr[ ], int y_arr[])
{
int i=0,j=0;
while (x_arr[i] != '0')
{
if (x_arr[i] == x_arr[i+1]) /*if current val. equals next val. -> jump dbl at x_arr*/
{
y_arr[j] = x_arr[i];
i+=2;
j++;
}
else
{
y_arr[j]=x_arr[i];
i++;
j++;
}
}
}
Please help me solve this problem...
thnx.
As a first hint, scanf returns the number of items read, so the condition (scanf("%d",&temp) ) != '0' will only be hit if you read 48 items (the ASCII value of 0). This isn't going to happen with that format specifier so that's why you've got the loop.