i dont have a c string ???
What does # mean when reading guitar tabs and why when reading them is there sometimes a 'C' string??
that would be a sharp note. one step higher than the note with out the sharp. :)
Reply:# is the sharp symbol
Reply:Ofcourse there's a C string...and # means sharp
Reply:a C sharp
Reply:Sharp Note. Usually up from a white key to a black key on a piano
Reply:I would assume that # means the same thing as it does in any musical notations - sharp, i.e. a half-tone up.
Reply:#: Sharp
♮: Natural
b: Flat
Reply:It's a sharped note; it means that you raise the pitch by a half-step (one fret, on a guitar).
Reply:No such thing as a C string so the letter is an indication of the chord. The # sign indicates a sharp note which is one fret higher than its usual position and the b sign indicates a flat which is one fret lower than the usual position. The Natural Sign is only used when a note has been sharped or flatted and needs to return to it's normal position within the same measure or between two bar lines OR when a note is sharped in the key signature and the notation requires it to be played in it's usual position. Of course these rules apply to musical notation on staff paper and not in the Tablature notation method.
You can browse through some of the offerings at ehow.com for "how to read guitar tablature" and see if you can pick up some new pointers.
Here is the link: http://www.ehow.com/Search.aspx?s=guitar...
Reply:sharp note
Tuesday, July 14, 2009
How can we store strings in an array in c?
that is, getting 3 names like john,david and deny using "for" loop then storing these names into an array using the same "for" loop.
simply getting and storing names will be done in the same "for" loop in any method like without using pointers.i've a doubt is there any method to getting and storing names in "for" loop without using pointers?if so please explain me........
How can we store strings in an array in c?
You may want to buy K%26amp;R's "The C Programming Language" which is the standard book for C programmers. They cover C strings. I'll point you to an online tutorial (http://www.cprogramming.com/tutorial/c/l... ).
There's two ways you can have a string. You can either have a pointer to a string literal, or have an array (either a pointer to one or the array object itself).
You may want to lookup the C standard library to see what you can use for input. (http://cppreference.com/stdio/index.html... The fgets function is particularly useful for this. If you look at the tutorial I pointed you to, they mention how to use it.
Reply:You can do it with a fixed-size array of char* s:
char* MyArray[ 3 ];
But no, there have to be pointers involved.
snow of june
simply getting and storing names will be done in the same "for" loop in any method like without using pointers.i've a doubt is there any method to getting and storing names in "for" loop without using pointers?if so please explain me........
How can we store strings in an array in c?
You may want to buy K%26amp;R's "The C Programming Language" which is the standard book for C programmers. They cover C strings. I'll point you to an online tutorial (http://www.cprogramming.com/tutorial/c/l... ).
There's two ways you can have a string. You can either have a pointer to a string literal, or have an array (either a pointer to one or the array object itself).
You may want to lookup the C standard library to see what you can use for input. (http://cppreference.com/stdio/index.html... The fgets function is particularly useful for this. If you look at the tutorial I pointed you to, they mention how to use it.
Reply:You can do it with a fixed-size array of char* s:
char* MyArray[ 3 ];
But no, there have to be pointers involved.
snow of june
C Passing a 3-D array of char strings to a function?
So I have a 3-D array of char strings -
char* val[256][256][256];
I need to pass a reference of this to a function, how do I do this?
If it was a single array I can pass it as follows
char* val[256]
func(val);
....
void func(char** a){}
However
void func(char**** a){}
doesnt work.
Any suggestions?
C Passing a 3-D array of char strings to a function?
int stPar (char a[5][5][5])
int stPar2 (char *a[256][5][5])
------
both of this work fine.
=====================================
Some thing to know more is in C
char *a;
char b[256][256];
a = b;
Works just as fine. The multiple index only indicates amount of memory. You can still pass as char * and then assign to deceleration as 3d array and it will work fine.
char* val[256][256][256];
I need to pass a reference of this to a function, how do I do this?
If it was a single array I can pass it as follows
char* val[256]
func(val);
....
void func(char** a){}
However
void func(char**** a){}
doesnt work.
Any suggestions?
C Passing a 3-D array of char strings to a function?
int stPar (char a[5][5][5])
int stPar2 (char *a[256][5][5])
------
both of this work fine.
=====================================
Some thing to know more is in C
char *a;
char b[256][256];
a = b;
Works just as fine. The multiple index only indicates amount of memory. You can still pass as char * and then assign to deceleration as 3d array and it will work fine.
Can any one of you solve this programm in C++, if P & T are strings with lengths R & S repectively?
And are stored as arrays with one charecter per element , find the indedx(location) of P in T??
Can any one of you solve this programm in C++, if P %26amp; T are strings with lengths R %26amp; S repectively?
Basically you have a couple of indexes
index_in_P
index_in_T
They both start at the beginning. You use a main for loop to walk through T.
if P[index_in_P] == T[index_in_T]
start seeing if the string matches all the way
for the rest of the letters in P (up to length R)
if P[index_in_P] == T[index_in_T+index_in_P]
you are still matching, good, increment index_in_P
else
they no longer match
index_in_T only increments when you are looking for a new start letter, it is the index you will return. If it is possible P is not in T, you need to make sure you don't go out of bounds in T.
Reply:Heres the C way:
int i, j;
for (i = 0; i %26lt; S - R; ++i)
{
for (j = 0; j %26lt; R; ++j)
{
if (P[j] != T[i+j])
break;
}
if (j == R) // got to the end of the loop successfully
break;
}
if (i == S - R) // got to the end without finding P in T
printf( "not found\n" );
else
printf( "found at %d\n", i );
In C++, I would use the string.find function:
size_type pos = T.find( S );
if (pos == npos) // npos is a constant
cout %26lt;%26lt; "not found" %26lt;%26lt; endl;
else
cout %26lt;%26lt; "found at " %26lt;%26lt; pos %26lt;%26lt; endl;
The more verbose way would be to use string.substr:
int pos;
for (pos = 0; pos %26lt; S - R; ++pos)
{
if (T.substr( pos, R ) == P)
break; // found it
}
if (pos == S - R)
cout %26lt;%26lt; "not found" %26lt;%26lt; endl;
else
cout %26lt;%26lt; "found at " %26lt;%26lt; pos %26lt;%26lt; endl;
And the most generic way using a C++ algorithm template:
char * pos = search( T, T+S, P, P+R );
if (pos == T+S)
cout %26lt;%26lt; "not found" %26lt;%26lt; endl;
else
cout %26lt;%26lt; "found at " %26lt;%26lt; (int)( pos - T ) %26lt;%26lt; endl;
Can any one of you solve this programm in C++, if P %26amp; T are strings with lengths R %26amp; S repectively?
Basically you have a couple of indexes
index_in_P
index_in_T
They both start at the beginning. You use a main for loop to walk through T.
if P[index_in_P] == T[index_in_T]
start seeing if the string matches all the way
for the rest of the letters in P (up to length R)
if P[index_in_P] == T[index_in_T+index_in_P]
you are still matching, good, increment index_in_P
else
they no longer match
index_in_T only increments when you are looking for a new start letter, it is the index you will return. If it is possible P is not in T, you need to make sure you don't go out of bounds in T.
Reply:Heres the C way:
int i, j;
for (i = 0; i %26lt; S - R; ++i)
{
for (j = 0; j %26lt; R; ++j)
{
if (P[j] != T[i+j])
break;
}
if (j == R) // got to the end of the loop successfully
break;
}
if (i == S - R) // got to the end without finding P in T
printf( "not found\n" );
else
printf( "found at %d\n", i );
In C++, I would use the string.find function:
size_type pos = T.find( S );
if (pos == npos) // npos is a constant
cout %26lt;%26lt; "not found" %26lt;%26lt; endl;
else
cout %26lt;%26lt; "found at " %26lt;%26lt; pos %26lt;%26lt; endl;
The more verbose way would be to use string.substr:
int pos;
for (pos = 0; pos %26lt; S - R; ++pos)
{
if (T.substr( pos, R ) == P)
break; // found it
}
if (pos == S - R)
cout %26lt;%26lt; "not found" %26lt;%26lt; endl;
else
cout %26lt;%26lt; "found at " %26lt;%26lt; pos %26lt;%26lt; endl;
And the most generic way using a C++ algorithm template:
char * pos = search( T, T+S, P, P+R );
if (pos == T+S)
cout %26lt;%26lt; "not found" %26lt;%26lt; endl;
else
cout %26lt;%26lt; "found at " %26lt;%26lt; (int)( pos - T ) %26lt;%26lt; endl;
What is the meaning of STRING in C?
What is C string? Can you please provide me good and exact answer. Thank you so much.
What is the meaning of STRING in C?
If you want to know how to work with it then search for some documentation. There you will find a lot of functions and operations.
Reply:An array is a data type that stores data in sequential memory slots.
A string is an array of characters.
Reply:A STRING, in essentially all programming languages, is a type of variable or constant which stores a series of alphanumreic characters, as opposed to an INTEGER which stores a number without a decimal and BOOLEAN which stores a 1 or a 0, a true/false value.
What is the meaning of STRING in C?
If you want to know how to work with it then search for some documentation. There you will find a lot of functions and operations.
Reply:An array is a data type that stores data in sequential memory slots.
A string is an array of characters.
Reply:A STRING, in essentially all programming languages, is a type of variable or constant which stores a series of alphanumreic characters, as opposed to an INTEGER which stores a number without a decimal and BOOLEAN which stores a 1 or a 0, a true/false value.
Plz solve this programm in C++, if P & T are strings with lengths R & S repectively and are stored as arrays?
with one charecter per element , find the indedx(location) of P in T??
Plz solve this programm in C++, if P %26amp; T are strings with lengths R %26amp; S repectively and are stored as arrays?
@P and @T ??
Reply:then u put in the oven to bake for half an hour lala x
sweet pea
Plz solve this programm in C++, if P %26amp; T are strings with lengths R %26amp; S repectively and are stored as arrays?
@P and @T ??
Reply:then u put in the oven to bake for half an hour lala x
sweet pea
Plz solve this programm in C++, if P & T are strings with lengths R & S repectively and are stored as arrays?
with one charecter per element , find the indedx(location) of P in T??
Plz solve this programm in C++, if P %26amp; T are strings with lengths R %26amp; S repectively and are stored as arrays?
loop through P %26amp; T until you find the character you're looking for, add 1 to "currentIndex" starting from zero each time through the loop, then return current index.
Reply:i dont think that there is any datatype say string however u can use char arrays oooh u mention arrays hope u r refering the same then
there are many cases to handle
for eg if R%26gt;S then u never find P in T however reverse is possible (hope u got my idea)
is P bigger than what T holds ??
and if all checks verify the possibility that P can be found in T then create an integer variable say index , set it to Zero and start searching char by char if the first char match incriment some pointer var say point=1 and increment till it matches the subsequent chars in T of P if it fails (means within point reaches the length of P it trace uncomman char then simply jump to index+point bcz u didnt find the match there and continue searching.....
if u reach the index %26gt; (T.length - P.length) (bcz after that u cant find P in T :-) means u dint find the string display sorry else if u finde in between make sure to toggle a flag variable so that after that loop u can trace that u find that string and u can also make a variable say counter =0 to check how many time u get that string.......
hope u get the best 4m me
in last plz rank my answer as best if u feel so
and comment plz :-) i m waiting
well as d below person use pointers and provide u d code i think it can be easily done by array indexing as i mention above and y he give d code it must let u first exercise yourself n then if cant implement then only code is given to you however, u r looking smart and i hope u can implement it by your own with my logics
good luck n great fun
http://www.geocities.com/ankur899/
Reply:Doing a homework assignment?
int (const char *p, int r, const char *t, int s)
{
int i;
for (i = 0; i %26lt;= (s-r); ++i)
{
if (strncmp(t+i, p, r) == 0) return (i);
}
return (-1); /* not found */
}
Of course, this is C, but C is a subset of C++, so this would work. I don't remember C++ has an exact replacement for C's strncpy() other than those that use the String class. But as you state your question, String probably isn't allowed.
To answer a comment by "d person above," please note that, in C, arrays and pointers are interchangeable in use. It's the declaration that's different. In particular, many people don't realize that *(a+i) is exactly the same as a[i] in C and C++. Exactly, as in by definition. That leads to the rather strange result that a[i] is exactly the same as i[a], even if 'i' is an int and 'a' is a pointer or array!
Anyway, certainly you can find harder ways to solve this problem by doing more of the work yourself. But this is the simplest way.
Plz solve this programm in C++, if P %26amp; T are strings with lengths R %26amp; S repectively and are stored as arrays?
loop through P %26amp; T until you find the character you're looking for, add 1 to "currentIndex" starting from zero each time through the loop, then return current index.
Reply:i dont think that there is any datatype say string however u can use char arrays oooh u mention arrays hope u r refering the same then
there are many cases to handle
for eg if R%26gt;S then u never find P in T however reverse is possible (hope u got my idea)
is P bigger than what T holds ??
and if all checks verify the possibility that P can be found in T then create an integer variable say index , set it to Zero and start searching char by char if the first char match incriment some pointer var say point=1 and increment till it matches the subsequent chars in T of P if it fails (means within point reaches the length of P it trace uncomman char then simply jump to index+point bcz u didnt find the match there and continue searching.....
if u reach the index %26gt; (T.length - P.length) (bcz after that u cant find P in T :-) means u dint find the string display sorry else if u finde in between make sure to toggle a flag variable so that after that loop u can trace that u find that string and u can also make a variable say counter =0 to check how many time u get that string.......
hope u get the best 4m me
in last plz rank my answer as best if u feel so
and comment plz :-) i m waiting
well as d below person use pointers and provide u d code i think it can be easily done by array indexing as i mention above and y he give d code it must let u first exercise yourself n then if cant implement then only code is given to you however, u r looking smart and i hope u can implement it by your own with my logics
good luck n great fun
http://www.geocities.com/ankur899/
Reply:Doing a homework assignment?
int (const char *p, int r, const char *t, int s)
{
int i;
for (i = 0; i %26lt;= (s-r); ++i)
{
if (strncmp(t+i, p, r) == 0) return (i);
}
return (-1); /* not found */
}
Of course, this is C, but C is a subset of C++, so this would work. I don't remember C++ has an exact replacement for C's strncpy() other than those that use the String class. But as you state your question, String probably isn't allowed.
To answer a comment by "d person above," please note that, in C, arrays and pointers are interchangeable in use. It's the declaration that's different. In particular, many people don't realize that *(a+i) is exactly the same as a[i] in C and C++. Exactly, as in by definition. That leads to the rather strange result that a[i] is exactly the same as i[a], even if 'i' is an int and 'a' is a pointer or array!
Anyway, certainly you can find harder ways to solve this problem by doing more of the work yourself. But this is the simplest way.
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