Repeating: Until you become expert at this I suggest you do this in two stages (using c and d separately). Suppose there are c digits after the decimal place where the digits are non-repeating, after which you get a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10c*(10d - 1), which is a string of d 9s followed by c 0s.For example 123.26159159… There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3. So the numerator is 12326159 – 12326 = 12313833 and the denominator is 99900 Therefore the fraction is 12313833/99900.Finally, you should check to see if the fraction can be simplified.
Until you become expert at this I suggest you do this in two stages (using c and d separately).Suppose there are c digits after the decimal place where the digits are non-repeating, after which you get a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10c*(10d - 1), which is a string of d 9s followed by c 0s.For example 123.26159159… There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3. So the numerator is 12326159 – 12326 = 12313833 and the denominator is 99900 Therefore the fraction is 12313833/99900.Check if the fraction can be simplified.
a + b = c To check if c is correct, subtract by either a or b , and you should get b and a respectively: a = c - b b = c - a
Until you become expert at this I suggest you do this in two stages (using c and d separately).Suppose there are c digits after the decimal place where the digits are non-repeating, and followed by a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10^c*(10^d - 1), which is a string of d 9s followed by c 0s.For example123.26159159… There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3.So the numerator is 12326159 – 12326 = 12313833and the denominator is 99900Therefore the fraction is 12313833/99900.
The following procedure is for converting a general repeating decimal, that is, one which does not start repeating straight away. Until you become expert at this I suggest you do this in two stages (using c and d separately). Suppose there are c digits after the decimal place where the digits are non-repeating, and then you get a repeating pattern of a string of d digits. Then the numerator is the old original string including one lot of the repeated digits minus the original string with none of the repeating digits. The denominator is 10^c*(10^d - 1), which is a string of d 9s followed by c 0s.Example123.26159159… There are 2 digits, "26", after the decimal point before the repeats kick in so c = 2, and the repeating string "159" is 3 digits long so d = 3.So the numerator of the fraction is 12326159 – 12326 = 12313833and the denominator is 99900Therefore the fraction is 12313833/99900.
A std::string is an object that encapsulates an array of type char whereas a C-style string is a primitive array with no members. A std::string is guaranteed to be null-terminated but a C-style string is not.
There is no C string on a violin- the strings are G, D, A and E. Perhaps you are thinking of a viola, which has a low C string, alongwith a G, D and an A string.
You can use "string" class in C++ for string operations or you may use c style string functions as well. #include <string> String class in C++ provides all basic function to operate on strings. you may details descriptin at http://www.cplusplus.com/reference/string/string/
yes, c-string is available for men..
Some stringed instruments contain a C-string, producing sound from the string creates the note C, so it is called C-string.
Use the Exception class and type converters.
Here is a page that will teach you about strings and pointers. www.juicystudio.com/tutorial/c/strings.asp Just saying String and Pointer don't actually aid us to describe what actually has to be answered .An Array is just repeatation of similar data types and String is as similar repeatation of continious characters in C. A String momentarily differs from a character array in terms of what the representations are for e.g. {'a','b','\0','i','s'}and{'a','b','\0','i','s','\0'}both are character arrays but in Ist "ab " is the string while in IInd "ab is" is as a whole is string for the string to be checked a ordidary way is to check the presence of the '\0' at last of its representation : Means :->check (Stringdef[i]!='\0')?HAS_MORE_CHAR:END_OF_STRING; for pointer validity:-> check (Ptr)?PTR_VALID:NOT_DEFINED; For more resurces just check any of the C material where the pointers are discussed and any misclearity mail me @ rupesh_joshi@sify.com,rupesh.joshi@gmail.com
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In C programming, a string doesn't have a specific return type as it's essentially an array of characters. So, if a function is returning a string, it should be declared to return a pointer to a char (char*), since a string in C is represented as an array of characters terminated by a null character ('\0').
C doesn't have String data-type, so don't use it.
The highest is the A string, then the D string, then the G string, and finally, the C string.
/*To check whether a string is palindrome*/includeincludevoid main () { int i,j,f=0; char a[10]; clrscr (); gets(a); for (i=0;a[i]!='\0';i++) { } i--; for (j=0;a[j]!='\0';j++,i--) { if (a[i]!=a[j]) f=1; } if (f==0) printf("string is palindrome"); else printf("string is not palindrome"); getch (); }