Program:
#include <iostream>
using namespace std;
int main() {
string str;
str[0] = 'a';
str[1] = 'b';
str[2] = 'c';
cout << str;
return 0;
}
Output:
No output.
If I replace cout << str; with cout << str[1], I get a proper output.
Output:
b
And if I change the data type of the variable to a character array, I get the desired output. (Replacing string str; with char str[5];
Output:
abc
Why is my program behaving like this? How do I alter my code to get the desired output without changing the data type?
Your program has undefined behavior.
string str;
creates an empty string. It has length 0.
You are trying to write to the first three elements of this string with
str[0] = 'a';
str[1] = 'b';
str[2] = 'c';
These do not exist. Indexing a std::string out-of-bounds causes undefined behavior.
You can add characters to a string with any of the following methods:
str += 'a';
str += "a";
str.push_back('a');
str.append("a");
or you can resize the string first to the intended length before you index into any of its elements:
str.resize(3);
As pointed out by #Ayxan in a comment under this answer, you are also missing #include<string>. Without it it is unspecified whether your program will compile since it uses std::string which is defined in <string>. It is unspecified whether including one standard library header will include another one if there isn't a specific exception. You should not rely on unspecified behavior that may break at any point.
In addition to walnut's answer, here's what's going on under the hood:
An std::string contains at least two members, a data pointer (type char*) and a size. The [] operator does not check the size, it only indexes into the memory behind the data pointer. Also, the [] operator does not modify the size member of the string.
The << operator that streams the string to cout, however, does interpret the size member. It does so to be able to output strings containing null characters. Since the size member is still zero, nothing is printed.
You may wonder why the memory access within str[0] even succeeds, after all, the string never had any reason to allocate any memory for its data yet. This is due to the fact that virtually all std::string implementations use the small-string-optimization: The std::string object itself is a bit larger than it needs to be, and the space at its end is used instead of an allocation on the heap unless the string becomes longer than that space. As such, the default constructor will just point the data pointer to that internal storage to have it initialized, and your memory accesses are directed to existing memory. This is why you don't get a SEGFAULT unless you access the string's data way out of bounds. Doesn't change the fact that already your expression str[0] is undefined behavior, though. The symptoms may appear benign, but the disease is fatal.
Related
std::string resize causes strings that appear to be equal to no longer be equal. It can appear to be misleading when I hover over the variable in my debugger and they appear to hold the same value.
I think it comes down to the fact that I expected the == operator to stop at the first null character but it keeps going till the end of the size. I'm sure this is working as intended but I was stuck on an issue caused by this for a while so I wanted to see why you would keep comparing characters even after the first null character. thanks!
int main(void)
{
std::string test1;
test1.resize(10);
test1[0] = 'a';
std::string test2 = "a";
//they are not equal
bool same = (test1 == test2);
return 0;
}
test1 is the string "a\0\0\0\0\0\0\0\0\0". test2 is the string "a". They are not equal.
std::string can contain null characters. Its length is not the distance to the first null character. It does also guarantee that the memory buffer containing the characters of the string ends with an additional null character 1 beyond its length.
If you don't intend for the string to be longer but just want the memory, use std::string::reserve. Note that you cannot access elements beyond the end with [] legally, but pushing back or whatever won't cause any new memory allocations until you pass the reserve limit.
This is the intended behavior of std::string. Unlike a c-string a std::string can have as many null characters as you want. For instance "this\0 is\0 a\0 legal\0 std::string\0" would be legal to have as the contents for a std::string. You have to build it like
std::string nulls_inside("this\0 is\0 a\0 legal\0 std::string\0", sizeof("this\0 is\0 a\0 legal\0 std::string\0");
but you can also insert null characters into an existing std::string. In your case you're comparing
"a\0\0\0\0\0\0\0\0\0\0"
against
"a\0"
so it fails.
I've been using:
string letters = THESAMELENGTH; // Assign for allocation purposes.
Reason being, if I:
string letters[THESAMELENGTH.length()];
I get a non constant expression complaint.
But if I:
string letters[12];
I'm at risk of needing to change every instance if the guide const string changes size.
But it seems foolish to assign a string when I won't use those entries, I only want my newly assigned string to be the same length as the previously assigned const string, then fill with different values.
How do you recommend I do this gracefully and safely?
You can
string letters(THESAMELENGTH.length(), ' '); // constructs the string with THESAMELENGTH.length() copies of character ' '
BTW: string letters[12]; doesn't mean the same as you expected. It declares a raw array of string containing 12 elements.
I only want my newly assigned string to be the same length as the previously assigned const string, then fill with different values.
Part of the reason the string class/type exists is so you don't have to worry about trying to manage its length. (The problem with arrays of char.)
If you have a const std::string tmp then you can't just assign anything to it after it has already been initialized. E.g.:
const std::string tmp = "A value"; // initialization
tmp = "Another value"; // compile error
How do you recommend I do this gracefully and safely?
If you really want to keep strings to a specific size, regardless of their contents, you could always resize your string variables. For example:
// in some constants.h file
const int MAX_STRING_LENGTH = 16;
// in other files
#include "constants.h"
// ...
std::string word = ... // some unknown string
word.resize(MAX_STRING_LENGTH);
Now your word string will have a length/size of MAX_STRING_LENGTH and anything beyond the end gets truncated.
This example is from C++ Reference
// resizing string
#include <iostream>
#include <string>
int main ()
{
std::string str ("I like to code in C");
std::cout << str << '\n';
unsigned sz = str.size();
str.resize (sz+2,'+');
std::cout << str << '\n';
str.resize (14);
std::cout << str << '\n';
return 0;
}
// program output
I like to code in C
I like to code in C++
I like to code
You can't just ask a string variable for its length at compile-time. By definition, it's impossible to know the value of a variable, or the state of any given program for that matter, while it's not running. This question only makes sense at run-time.
Others have mentioned this, but there seems to be an issue with your understanding of string letters[12];. That gives you an array of string types, i.e. you get space for 12 full strings (e.g. words/sentences/etc), not just letters.
In other words, you could do:
for(size_t i = 0; i < letters.size(); ++i)
letters[i] = "Hello, world!";
So your letters variable should be renamed to something more accurate (e.g. words).
If you really want letters (e.g. the full alphabet on a single string), you could do something like this:
// constants.h
const std::string ALPHABET_LC = "abc...z";
const std::string ALPHABET_UC = "ABC...Z";
const int LETTER_A = 0;
const int LETTER_B = 1;
// ...
// main.cpp, etc.
char a = ALPHABET_LC[LETTER_A];
char B = ALPHABET_UC[LETTER_B];
// ...
It all depends on what you need to do, but this might be a good alternative.
Disclaimer: Note that it's not really my recommendation that you do this. You should let strings manage their own length. For example, if the string value is actually shorter than your limit, you're causing your variable to use more space/memory than needed, and if it's longer, you're still truncating it. Neither side-effect is good, IMHO.
The first thing you need to do is understand the difference between a string length and an array dimension.
std::string letters = "Hello";
creates a single string that contains the characters from "Hello", and has length 5.
In comparison
std::string letters[5];
creates an array of five distinct default-constructed objects of type std::string. It doesn't create a single string of 5 characters. The reason for the non-constant complaint when doing
std::string letters[THESAMELENGTH.length()];
is that construction of arrays in standard C++ is required to use a length known to the compiler, whereas the length of a std::string is determined at run time.
If you have a string, and what to create another string of the same length, you can do something like
std::string another_string(letters.length(), 'A');
which will create a single string containing the required number of letters 'A'.
It is largely pointless to do what you are seeking as a std::string can dynamically change its length anyway, as needed. There is also nothing stopping a std::string from allocating more than it needs (e.g. to make provision for multiple increases in its length).
To change the contents of a string in a function such that it reflects in the main function we need to accept the string as reference as indicated below.
Changing contents of a std::string with a function
But in the above code we are changing the size of string also(i.e, more than what it can hold), so why is the program not crashing ?
Program to convert decimal to binary, mind it, the code is not complete and I am just testing the 1st part of the code.
void dectobin(string & bin, int n)
{
int i=0;
while(n!=0)
{
bin[i++]= (n % 2) + '0';
n = n / 2;
}
cout << i << endl;
cout << bin.size() << endl;
cout << bin << endl;
}
int main()
{
string s = "1";
dectobin(s,55);
cout << s << endl;
return 0;
}
O/p: 6 1 1 and the program crashes in codeblocks. While the above code in the link works perfectly fine.
It only outputs the correct result, when i initialize the string in main with 6 characters(i.e, length of the number after it converts from decimal to binary).
http://www.cplusplus.com/reference/string/string/capacity/
Notice that this capacity does not suppose a limit on the length of the string. When this capacity is exhausted and more is needed, it is automatically expanded by the object (reallocating it storage space). The theoretical limit on the length of a string is given by member max_size
If the string resizes itself automatically then why do we need the resize function and then why is my decimal to binary code not working?
Your premise is wrong. You are thinking 1) if I access a string out of bound then my program will crash, 2) my program doesn't crash therefore I can't be accessing a string out of bounds, 3) therefore my apparently out of bounds string accesses must actually resize the string.
1) is incorrect. Accessing a string out of bounds results in undefined behaviour. This is means exactly what it says. Your program might crash but it might not, it's behaviour is undefined.
And it's a fact that accessing a string never changes it's size, that's why we have the resize function (and push_back etc.).
We must get questions like yours several times a week. Undefined behaviour is clearly a concept that newbies find surprising.
Check this link about std::string:
char& operator[] (size_t pos);
const char& operator[] (size_t pos) const;
If pos is not greater than the string length, the function never
throws exceptions (no-throw guarantee). Otherwise, it causes
undefined behavior.
In your while loop you are accessing the bin string with index that is greater than bin.size()
You aren't changing the size of the string anywhere. If the string you pass into the function is of length one and you access it at indices larger than 0, i.e., at bin[1], bin[2], you are not modifying the string but some other memory locations after the string - there might be something else stored there. Corrupting memory in this way does not necessarily directly lead to a crash or an exception. It will once you access those memory locations later on in your program.
Accepting a reference to a string makes it possible to change instances of strings from the calling code inside the called code:
void access(std::string & str) {
// str is the same instance as the function
// is called with.
// without the reference, a copy would be made,
// then there would be two distinct instances
}
// ...
std::string input = "test";
access(input);
// ...
So any function or operator that is called on a reference is effectively called on the referenced instance.
When, similar to your linked question, the code
str = " new contents";
is inside of the body of the access function, then operator= of the input instance is called.
This (copy assignment) operator is discarding the previous contents of the string, and then copying the characters of its argument into newly allocated storage, whose needed length is determined before.
On the other hand, when you have code like
str[1] = 'a';
inside the access function, then this calls operator[] on the input instance. This operator is only providing access to the underlying storage of the string, and not doing any resizing.
So your issues aren't related to the reference, but to misusing the index operator[]:
Calling that operator with an argument that's not less than the strings size/length leads to undefined behaviour.
To fix that, you could resize the string manually before using the index operator.
As a side note: IMO you should try to write your function in a more functional way:
std::string toOct(std::string const &);
That is, instead of modifying the oases string, create a new one.
The bounds of the string are limited by its current content. That is why when you initialise the string with 6 characters you will stay inside bounds for conversion of 55 to binary and program runs without error.
The automatic expansion feature of strings can be utilised using
std::string::operator+=
to append characters at the end of current string. Changed code snippet will look like this:
void dectobin(string & bin, int n){
//...
bin += (n % 2) + '0';
//...
}
Plus you don't need to initialise the original string in main() and your program should now run for arbitrary decimals as well.
int main(){
//...
string s;
dectobin(s,55);
//...
}
I am taking a line of input which is separated by a space and trying to read the data into two integer variables.
for instance: "0 1" should give child1 == 0, child2 == 1.
The code I'm using is as follows:
int separator = input.find(' ');
const char* child1_str = input.substr(0, separator).c_str(); // Everything is as expected here.
const char* child2_str = input.substr(
separator+1, //Start with the next char after the separator
input.length()-(separator+1) // And work to the end of the input string.
).c_str(); // But now child1_str is showing the same location in memory as child2_str!
int child1 = atoi(child1_str);
int child2 = atoi(child2_str); // and thus are both of these getting assigned the integer '1'.
// do work
What's happening is perplexing me to no end. I'm monitoring the sequence with the Eclipse debugger (gdb). When the function starts, child1_str and child2_str are shown to have different memory locations (as they should). After splitting the string at separator and getting the first value, child1_str holds '0' as expected.
However, the next line, which assigns a value to child2_str not only assigns the correct value to child2_str, but also overwrites child1_str. I don't even mean the character value is overwritten, I mean that the debugger shows child1_str and child2_str to share the same location in memory.
What the what?
1) Yes, I'll be happy to listen to other suggestions to convert a string to an int -- this was how I learned to do it a long time ago, and I've never had a problem with it, so never needed to change, however:
2) Even if there's a better way to perform the conversion, I would still like to know what's going on here! This is my ultimate question. So even if you come up with a better algorithm, the selected answer will be the one that helps me understand why my algorithm fails.
3) Yes, I know that std::string is C++ and const char* is standard C. atoi requires a c string. I'm tagging this as C++ because the input will absolutely be coming as a std::string from the framework I am using.
First, the superior solutions.
In C++11 you can use the newfangled std::stoi function:
int child1 = std::stoi(input.substr(0, separator));
Failing that, you can use boost::lexical_cast:
int child1 = boost::lexical_cast<int>(input.substr(0, separator));
Now, an explanation.
input.substr(0, separator) creates a temporary std::string object that dies at the semicolon. Calling c_str() on that temporary object gives you a pointer that is only valid as long as the temporary lives. This means that, on the next line, the pointer is already invalid. Dereferencing that pointer has undefined behaviour. Then weird things happens, as is often the case with undefined behaviour.
The value returned by c_str() is invalid after the string is destructed. So when you run this line:
const char* child1_str = input.substr(0, separator).c_str();
The substr function returns a temporary string. After the line is run, this temporary string is destructed and the child1_str pointer becomes invalid. Accessing that pointer results in undefined behavior.
What you should do is assign the result of substr to a local std::string variable. Then you can call c_str() on that variable, and the result will be valid until the variable is destructed (at the end of the block).
Others have already pointed out the problem with your current code. Here's how I'd do the conversion:
std::istringstream buffer(input);
buffer >> child1 >> child2;
Much simpler and more straightforward, not to mention considerably more flexible (e.g., it'll continue to work even if the input has a tab or two spaces between the numbers).
input.substr returns a temporary std::string. Since you are not saving it anywhere, it gets destroyed. Anything that happens afterwards depends solely on your luck.
I recommend using an istringstream.
could you say me what is the mistake in my following code?
char* line="";
printf("Write the line.\n");
scanf("%s",line);
printf(line,"\n");
I'm trying to get a line as an input from the console.But everytime while using "scanf" the program crashes. I don't want to use any std, I totally want to avoid using cin or cout. I'm just trying to learn how to tak a full line as an input using scanf().
Thank you.
You need to allocate the space for the input string as sscanf() cannot do that itself:
char line[1024];
printf("Write the line.\n");
scanf("%s",line);
printf(line,"\n");
However this is dangerous as it's possible to overflow the buffer and is therefore a security concern. Use std::string instead:
std::string line;
std::cout << "Write the line." << std::endl;
std::cin >> line;
std::cout << line << std::endl;
or:
std::getline (std::cin, line);
Space not allocated for line You need to do something like
char *line = malloc();
or
Char line[SOME_VALUE];
Currently line is a poor pointer pointing at a string literal. And overwriting a string literal can result in undefined behaviour.
scanf() doesn't match lines.
%s matches a single word.
#include <stdio.h>
int main() {
char word[101];
scanf("%100s", word);
printf("word <%s>\n", word);
return 0;
}
input:
this is a test
output:
word <this>
to match the line use %100[^\n"] which means 100 char's that aren't newline.
#include <stdio.h>
int main() {
char word[101];
scanf("%100[^\n]", word);
printf("word <%s>\n", word);
return 0;
}
You are trying to change a string literal, which in C results in Undefined behavior, and in C++ is trying to write into a const memory.
To overcome it, you might want to allocate a char[] and assign it to line - or if it is C++ - use std::string and avoid a lot of pain.
You should allocate enough memory for line:
char line[100];
for example.
The %s conversion specifier in a scanf call expects its corresponding argument to point to a writable buffer of type char [N] where N is large enough to hold the input.
You've initialized line to point to the string literal "". There are two problems with this. First is that attempting to modify the contents of a string literal results in undefined behavior. The language definition doesn't specify how string literals are stored; it only specifies their lifetime and visibility, and some platforms stick them in a read-only memory segment while others put them in a writable data segment. Therefore, attempting to modify the contents of a string literal on one platform may crash outright due to an access violation, while the same thing on another platform may work fine. The language definition doesn't mandate what should happen when you try to modify a string literal; in fact, it explicitly leaves that behavior undefined, so that the compiler is free to handle the situation any way it wants to. In general, it's best to always assume that string literals are unwritable.
The other problem is that the array containing the string literal is only sized to hold 1 character, the 0 terminator. Remember that C-style strings are stored as simple arrays of char, and arrays don't automatically grow when you add more characters.
You will need to either declared line as an array of char or allocate the memory dynamically:
char line[MAX_INPUT_LEN];
or
char *line = malloc(INITIAL_INPUT_LEN);
The virtue of allocating the memory dynamically is that you can resize the buffer as necessary.
For safety's sake, you should specify the maximum number of characters to read; if your buffer is sized to hold 21 characters, then write your scanf call as
scanf("%20s", line);
If there are more characters in the input stream than what line can hold, scanf will write those extra characters to the memory following line, potentially clobbering something important. Buffer overflows are a common malware exploit and should be avoided.
Also, %s won't get you the full line; it'll read up to the next whitespace character, even with the field width specifier. You'll either need to use a different conversion specifier like %[^\n] or use fgets() instead.
The pointer line which is supposed to point to the start of the character array that will hold the string read is actually pointing to a string literal (empty string) whose contents are not modifiable. This leads to an undefined behaviour manifested as a crash in your case.
To fix this change the definition to:
char line[MAX]; // set suitable value for MAX
and read atmost MAX-1 number of characters into line.
Change:
char* line="";
to
char line[max_length_of_line_you_expect];
scanf is trying to write more characters than the reserved by line. Try reserving more characters than the line you expect, as been pointed out by the answers above.