if statement obj-c zero input - objective-c

here is code.
int main(int argc, const char * argv[])
{
#autoreleasepool {
int x,y;
//BOOL divsibleYESOrNo;
NSLog(#"enter two number for test it\n");
scanf("%i %i",&x,&y);
if ( x%y ==0 ) {
NSLog(#"YES,it can be");
}
else if (x%y !=0) {
NSLog(#"no there cant.");
}
else
NSLog(#"zero is not allow");
}
return 0;
}
this code can not detect if the user input two zero.
how can I modify this code that can detect the input values are zero?
Thanks

Anything % 0 is undefined, so you'll need to add a check for y == 0 before your other if statements. It has to be the first if statement in order to make sure it gets evaluated; otherwise one of your other statements will catch it erroneously first.
if ( y == 0 ) {
NSLog(#"zero is not allowed.");
}
else if ( x%y == 0 ) {
NSLog(#"YES,it can be");
}
else {
NSLog(#"no there cant.");
}

Related

struct entry: Manual command line access?

how does this code work: more precisely the line that passes the struct into the command line agc and argv ? what is it doing? Is there another way to do this?
#include <Foundation/Foundation.h>
int main (int argc, char * argv[]) {
struct entry dictionary[100] =
{ { "aardvark", "a burrowing African mammal" } ,
{ "abyss", "a bottomless pit' } ,
{ "acumen", "mentally sharp; keen"
{ "addle", "to become confused" } ,
{ "aerie", "a high nest" },
{ "affix","to append, attach" },
{ "agar"," a jelly made from seaweed" },
{ "ahoy"," a nautical call of greeting" },
{ "aigrette", "an ornamental cluster of feathers" } ,
{ "ajar", "partially opened" } } ;
int entries = 10;
int entryNumber;
int lookup (struct entry dictionary [], char search[],
int entries);
if ( argc != 2 ) {
NSLog (#"No word typed on the command line.");
return (1); }
entryNumber = lookup (dictionary, argv[1], entries);
if ( entryNumber != -1 )
NSLog (#"%s", dictionary[entryNumber].definition);
else
NSLog (#"Sorry, %s is not in my dictionary.", argv[1]);
return (0); }

How to check that two format strings are compatible?

Examples:
"Something %d" and "Something else %d" // Compatible
"Something %d" and "Something else %f" // Not Compatible
"Something %d" and "Something %d else %d" // Not Compatible
"Something %d and %f" and "Something %2$f and %1$d" // Compatible
I figured there should be some C function for this, but I'm not getting any relevant search results. I mean the compiler is checking that the format string and the arguments match, so the code for checking this is already written. The only question is how I can call it.
I'm using Objective-C, so if there is an Objective-C specific solution that's fine too.
Checking if 2 printf() format strings are compatible is an exercise in format parsing.
C, at least, has no standard run-time compare function such as:
int format_cmp(const char *f1, const char *f2); // Does not exist
Formats like "%d %f" and "%i %e" are obviously compatible in that both expect an int and float/double. Note: float are promoted to double as short and signed char are promoted to int.
Formats "%*.*f" and "%i %d %e" are compatible, but not obvious: both expect an int,int and float/double.
Formats "%hhd" and "%d" both expect an int, even though the first will have it values cast to signed char before printing.
Formats "%d" and "%u" are not compatible. Even though many systems will behaved as hoped. Note: Typically char will promote to int.
Formats "%d" and "%ld" are not strictly compatible. On a 32-bit system there are equivalent, but not in general. Of course code can be altered to accommodate this. OTOH "%lf" and "%f" are compatible due to the usual argument promotions of float to double.
Formats "%lu" and "%zu" may be compatible, but that depends on the implementation of unsigned long and size_t. Additions to code could allow this or related equivalences.
Some combinations of modifiers and specifiers are not defined like "%zp". The following does not dis-allow such esoteric combinations - but does compare them.
Modifiers like "$" are extensions to standard C and are not implemented in the following.
The compatibility test for printf() differs from scanf().
#include <ctype.h>
#include <limits.h>
#include <stdio.h>
#include <string.h>
typedef enum {
type_none,
type_int,
type_unsigned,
type_float,
type_charpointer,
type_voidpointer,
type_intpointer,
type_unknown,
type_type_N = 0xFFFFFF
} type_type;
typedef struct {
const char *format;
int int_queue;
type_type type;
} format_T;
static void format_init(format_T *state, const char *format);
static type_type format_get(format_T *state);
static void format_next(format_T *state);
void format_init(format_T *state, const char *format) {
state->format = format;
state->int_queue = 0;
state->type = type_none;
format_next(state);
}
type_type format_get(format_T *state) {
if (state->int_queue > 0) {
return type_int;
}
return state->type;
}
const char *seek_flag(const char *format) {
while (strchr("-+ #0", *format) != NULL)
format++;
return format;
}
const char *seek_width(const char *format, int *int_queue) {
*int_queue = 0;
if (*format == '*') {
format++;
(*int_queue)++;
} else {
while (isdigit((unsigned char ) *format))
format++;
}
if (*format == '.') {
if (*format == '*') {
format++;
(*int_queue)++;
} else {
while (isdigit((unsigned char ) *format))
format++;
}
}
return format;
}
const char *seek_mod(const char *format, int *mod) {
*mod = 0;
if (format[0] == 'h' && format[1] == 'h') {
format += 2;
} else if (format[0] == 'l' && format[1] == 'l') {
*mod = ('l' << CHAR_BIT) + 'l';
format += 2;
} else if (strchr("ljztL", *format)) {
*mod = *format;
format++;
} else if (strchr("h", *format)) {
format++;
}
return format;
}
const char *seek_specifier(const char *format, int mod, type_type *type) {
if (strchr("di", *format)) {
*type = type_int;
format++;
} else if (strchr("ouxX", *format)) {
*type = type_unsigned;
format++;
} else if (strchr("fFeEgGaA", *format)) {
if (mod == 'l') mod = 0;
*type = type_float;
format++;
} else if (strchr("c", *format)) {
*type = type_int;
format++;
} else if (strchr("s", *format)) {
*type = type_charpointer;
format++;
} else if (strchr("p", *format)) {
*type = type_voidpointer;
format++;
} else if (strchr("n", *format)) {
*type = type_intpointer;
format++;
} else {
*type = type_unknown;
exit(1);
}
*type |= mod << CHAR_BIT; // Bring in modifier
return format;
}
void format_next(format_T *state) {
if (state->int_queue > 0) {
state->int_queue--;
return;
}
while (*state->format) {
if (state->format[0] == '%') {
state->format++;
if (state->format[0] == '%') {
state->format++;
continue;
}
state->format = seek_flag(state->format);
state->format = seek_width(state->format, &state->int_queue);
int mod;
state->format = seek_mod(state->format, &mod);
state->format = seek_specifier(state->format, mod, &state->type);
return;
} else {
state->format++;
}
}
state->type = type_none;
}
// 0 Compatible
// 1 Not Compatible
// 2 Not Comparable
int format_cmp(const char *f1, const char *f2) {
format_T state1;
format_init(&state1, f1);
format_T state2;
format_init(&state2, f2);
while (format_get(&state1) == format_get(&state2)) {
if (format_get(&state1) == type_none)
return 0;
if (format_get(&state1) == type_unknown)
return 2;
format_next(&state1);
format_next(&state2);
}
if (format_get(&state1) == type_unknown)
return 2;
if (format_get(&state2) == type_unknown)
return 2;
return 1;
}
Note: only minimal testing done. Lots of additional considerations could be added.
Known shortcomings: hh,h,l,ll,j,z,t modifiers with n. l with s,c.
[Edit]
OP comments about security concerns. This changes the nature of the post and the compare from an equality one to a security one. I'd imagine that one of the patterns (A) would be a reference pattern and the next (B) would be the test. The test would be "is B at least as secure as A?". Example A = "%.20s" and B1 = "%.19s", B2 = "%.20s", B3 = "%.21s". B1 and B2 both pass the security test as they do not extract more the 20 char. B3 is a problem as it goes pass the reference limit of 20 char. Further any non-width qualified with %s %[ %c is a security problem - in the reference or test pattern. This answer's code does not address this issue.
As mentioned, code does not yet handle modifiers with "%n".
[2018 edit]
Concerning "Formats "%d" and "%u" are not compatible.": This is for values to be printed in general. For values in the [0..INT_MAX] range, either format may work per C11dr ยง6.5.2.2 6.
My understanding of what you want, is that, you basically want a method which can look at two strings and detect if they both have the same types of values in them. Or something a long those lines.... If so, then try this (or something along the lines of this):
-(int)checkCompatible:(NSString *)string_1 :(NSString *)string_2 {
// Separate the string into single elements.
NSArray *stringArray_1 = [string_1 componentsSeparatedByString:#" "];
NSArray *stringArray_2 = [string_2 componentsSeparatedByString:#" "];
// Store only the numbers for comparison in a new array.
NSMutableArray *numbers_1 = [[NSMutableArray alloc] init];
NSMutableArray *numbers_2 = [[NSMutableArray alloc] init];
// Make sure the for loop below, runs for the appropriate
// number of cycles depending on which array is bigger.
int loopMax = 0;
if ([stringArray_1 count] > [stringArray_2 count]) {
loopMax = (int)[stringArray_1 count];
}
else {
loopMax = (int)[stringArray_2 count];
}
// Now go through the stringArray's and store only the
// numbers in the mutable array's. This will be used
// during the comparison stage.
for (int loop = 0; loop < loopMax; loop++) {
NSCharacterSet *notDigits = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
if (loop < [stringArray_1 count]) {
if ([[stringArray_1 objectAtindex:loop] rangeOfCharacterFromSet:notDigits].location == NSNotFound) {
// String consists only of the digits 0 through 9.
[numbers_1 addObject:[stringArray_1 objectAtindex:loop]];
}
}
if (loop < [stringArray_2 count]) {
if ([[stringArray_2 objectAtindex:loop] rangeOfCharacterFromSet:notDigits].location == NSNotFound) {
// String consists only of the digits 0 through 9.
[numbers_2 addObject:[stringArray_2 objectAtindex:loop]];
}
}
}
// Now look through the mutable array's
// and perform the type comparison,.
if ([numbers_1 count] != [numbers_2 count]) {
// One of the two strings has more numbers
// than the other, so they are NOT compatible.
return 1;
}
else {
// Both string have the same number of numbers
// numbers so lets go through them to make
// sure the numbers are of the same type.
for (int loop = 0; loop < [numbers_1 count]; loop++) {
// Check to see if the number in the current array index
// is a float or an integer. All the numbers in the array have
// to be the SAME type, in order for the strings to be compatible.
BOOL check_float_1 = [[NSScanner scannerWithString:[numbers_1 objectAtindex:loop]] scanFloat:nil];
BOOL check_int_1 = [[NSScanner scannerWithString:[numbers_1 objectAtindex:loop]] scanInt:nil];
BOOL check_float_2 = [[NSScanner scannerWithString:[numbers_2 objectAtindex:loop]] scanFloat:nil];
BOOL check_int_2 = [[NSScanner scannerWithString:[numbers_2 objectAtindex:loop]] scanInt:nil];
if (check_float_1 == YES) {
if (check_float_2 == NO) {
return 1;
}
}
else if (check_int_1 == YES) {
if (check_int_2 == NO) {
return 1;
}
}
else {
// Error of some sort......
return 1;
}
}
// All the numbers in the strings are of the same
// type (otherwise we would NOT have reached
// this point). Therefore the strings are compatible.
return 0;
}
}

Numbers into words issue C

I am trying to write a program that will take an integer input, and then convert it to words. for example: 123, one two three. Also -3908, negative three nine zero eight.
My code works 90% of the time, the only issue coming along when i am putting one or more zeros on the end of the integer. eg. 70800 will come up as seven zero eight. It completely misses the end zeros. I understand why that is happening but does anybody know if there is a way around it.
PS(i am not allowed as a part of this task to accept the input as a string and split it into an array, so it would be best if the answer is based off this code).
int main(int argc, const char * argv[])
{
#autoreleasepool {
float abNumber;
int i = 0;
float number;
float result;
float firstNumber;
printf("type a number: ");
scanf("%f", &firstNumber);
abNumber = abs(firstNumber);
if (firstNumber < 0) {
printf("negative ");
}
number = abNumber;
while (number >= 10) {
number = number / 10;
i++;
}
do {
float countNumber = abNumber;
float power = powf(10, -i);
float powerNo2 = powf(10, i);
countNumber = countNumber * power;
result = floorf(countNumber);
if (result == 9){
printf("nine ");
}
if (result == 8){
printf("eight ");
}
if (result == 7){
printf("seven ");
}
if (result == 6){
printf("six ");
}
if (result == 5){
printf("five ");
}
if (result == 4){
printf("four ");
}
if (result == 3){
printf("three ");
}
if (result == 2){
printf("two ");
}
if (result == 1){
printf("one ");
}
if (result == 0){
printf("zero ");
}
while (abNumber > powerNo2) {
abNumber = abNumber - powerNo2;
}
i--;
} while (i >= 0);
}
return 0;
}
The main error seems to be that
while (abNumber > powerNo2) {
should be
while (abNumber >= powerNo2) {
But I would recommend not to use floating point arithmetic at all, to avoid
possible rounding errors. The same can be achieved with simple integer arithmetic
(I have omitted the "negative case" for simplicity):
int number;
printf("type a number: ");
scanf("%d", &number);
// Determine highest power of 10 that is <= the given number:
int power = 1;
while (10 * power <= number) {
power *= 10;
}
// Extract each digit:
while (power > 0) {
int digit = (number / power) % 10;
/*
* Use switch/case to print 'digit' as a string ...
*/
power /= 10;
}
I would go for recursive solution, like that
int print(int num)
{
if( num )
{
int mod = num%10;
print(num/10);
switch(mod)
{
case 0:printf(" zero");break;
case 1:printf(" one");break;
case 2:printf(" two");break;
case 3:printf(" three");break;
}
}
return 0;
}
Recursivy divide the number untill nothing left of it, on the way back print the mod.
Why don't you just input the number as a string then loop through each character:
Exemple: http://ideone.com/E8QspN
Input:
-12003200
Output:
negative one two zero zero tree two zero zero
Code:
#include <stdio.h>
int main(int argc, char *argv[])
{
char input[25];
scanf("%s", input);
int i = 0;
while (input[i] != '\0') {
switch(input[i]) {
case '-' :
printf("negative");
break;
case '0' :
printf("zero");
break;
case '1' :
printf("one");
break;
case '2' :
printf("two");
break;
case '3' :
printf("tree");
break;
case '4' :
printf("four");
break;
case '5' :
printf("five");
break;
case '6' :
printf("six");
break;
case '7' :
printf("seven");
break;
case '8' :
printf("eight");
break;
case '9' :
printf("nine");
break;
default :
break;
}
printf(" ");
i++;
}
return 0;
}
Consider the following:
Code
#include <stdio.h>
#include <stdlib.h>
const char *numbers[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
void printNum(int num);
int main(void)
{
int num;
printf("Enter a number: ");
scanf("%u", &num);
printNum(num);
printf("\n");
return 0;
}
void printNum(int num)
{
int absNum = abs(num);
if(absNum > 9)
printNum(num / 10);
if((absNum < 10) && (num < 0))
printf("negative");
printf(" %s", numbers[absNum % 10]);
}
Example Output
Enter a number: 2582
two five eight two
Enter a number: -943
negative nine four three
Enter a number: 1000
one zero zero zero
Enter a number: -1000
negative one zero zero zero
Logic
Get an integer from user.
Send to recursive function.
Keep recursing until the least significant digit is left. This is to print in the correct order.
Print digit as a string by using a lookup table.
Keep someone a high five.
To Do
Error checking

Expected expression error. C

I am trying to write a code that will take two intergers, and will list all numbers lower than both of them except for those that are a factor of either if the two numbers inputted. At some point in my code though (see below) i am getting an error saying expected expression. I am a beginner so if you could explain this too me as simply as possible.
int main(int argc, const char * argv[])
{
#autoreleasepool {
int firstInterger;
int secondInterger;
int i;
printf("Please enter the first interger: ");
scanf("%i", &firstInterger);
printf("Please enter the second interger:");
scanf("%i", &secondInterger);
for (i = 0; i < firstInterger && i < secondInterger; i++) {
if ((firstInterger % i !== 0) && (secondInterger % i !== 0)) { //ERROR HERE!
printf("%i", i);
}
}
}
return 0;
}
You have twice the same error in that line.
You should replace !== with != or ==

Use of blocks in Objective-C

const char *sentence = "He was not in the cab at the time.";
printf("\"%s\" has %d spaces\n", sentence, (int) ^ {
int i = 0;
int countSpaces = 0;
while (sentence[i] != '\0') {
if (sentence[i] == 0x20) {
countSpaces++;
}
i++;
}
return countSpaces;
});
This code simply counts the white space in a string, but for some reason it says 1606416608 spaces rather than 8. I'm not exactly sure what is going wrong, so thanks for any help!
You're passing the actual block to printf, not the result of the block. Instead, try
const char *sentence = "He was not in the cab at the time.";
printf("\"%s\" has %d spaces\n", sentence, (int) ^ {
int i = 0;
int countSpaces = 0;
while (sentence[i] != '\0') {
if (sentence[i] == 0x20) {
countSpaces++;
}
i++;
}
return countSpaces;
}()); // <-- note the extra parentheses here, indicating that you're calling the block