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C++ String Causes Segmentation Fault

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Not the answer you're looking for? Replace the statement with something like: Code: std::string *goodQstring = new std::string("sdfsdf"); P.S. Also, many times a function requires that an address (corresponding to a parameter of pointer type) be sent to it as an argument (as is true of many of the Numerical Visit the following links: Site Howto | Site FAQ | Sitemap | Register Now If you have any problems with the registration process or your account login, please contact us. Check This Out

I tried that and it resulted in a strange error named "Illegal instruction". As for decade of experience - while my comments won't be useful for you, they might be useful for somebody who finds this thread later. The literal string is not writable so when you do: str[0] = 'z'; you get a seg fault. For example: int * ptr; int variable; ptr = &variable; Or, equivalently, int variable; int *ptr = &variable; Other common ways include assigning the pointer the address of memory http://stackoverflow.com/questions/164194/why-do-i-get-a-segmentation-fault-when-writing-to-a-string-initialized-with-cha

C++ String Causes Segmentation Fault

Join them; it only takes a minute: Sign up String - Segmentation Fault up vote 0 down vote favorite I have a problem with this code. Last edited by dwhitney67; 04-29-2011 at 05:39 AM. 1 members found this post helpful. How can I master those concepts to get rid of segmentation...How can I prevent runtime errors in my program?Why do I get Segmentation fault error when executing my code?I don't get scanf("%s", name_1); You haven't allocated any space for name_1 to point to.

Browse other questions tagged c segfault windows . If you attempt to use reinterpret_cast<> or static_cast<>, then a warning would be generated. What does "Game of the Year" actually mean? C Segmentation Fault 11 It is considered poor programming practice.

int a[2]={0,1}; printf("%d\n", a[3]); // this will work. C++ String Segmentation Fault It attempts to modify a string literal, which is undefined behavior according to the ANSI C standard. Equivalent for "Crowd" in the context of machines Where's the 0xBEEF? http://stackoverflow.com/questions/943312/segmentation-fault-char-pointer Are illegal immigrants more likely to commit crimes?

Most compilers will not catch this at compile time, and instead compile this to executable code that will crash: int main(void) { char *s = "hello world"; *s = 'H'; } How To Remove Segmentation Fault In C I am using a gcc compiler and when i compile and execute this code i am getting a seg fault. I asked a similar question here but I'm not sure if this is because I have two copies of num. If we are freeing memory twice !free(ptr); free(ptr); 4.

C++ String Segmentation Fault

I thing segmentation faults are specific to Linux/UNIX platforms. –M.S. http://web.mit.edu/10.001/Web/Tips/tips_on_segmentation.html more hot questions question feed default about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation C++ String Causes Segmentation Fault The first allocates 100 bytes and sets num to point to those bytes. What Is A Segmentation Fault In C This declaration is identical to char s[] = { 'a', 'b', 'c', '\0' }, t[] = { 'a', 'b', 'c' }; The contents of the arrays are modifiable.

How to apply a constant function to a vector of values? his comment is here Would it be ok to eat rice using a spoon in front of Westerners? How come Ferengi starships work? As for avoiding direct calls to "new", I'd would like to see an example where one could allocate an object without using "new" (or "malloc" family of functions). C Segmentation Fault Core Dumped

Please explain why it's seg-faulting. If so, why is it allowed? Second is an array with initialized value, so it can be modified. this contact form For more advanced trainees it can be a desktop reference, and a collection of the base knowledge needed to proceed with system and network administration.

The key idea, however, is that all segment faults are caused by accessing memory that your program doesn't own.37.1k Views · View Upvotes · Answer requested by Gami Nipulkumar Anhad Jai How To Debug Segmentation Fault I thing segmentation faults are specific to Linux/UNIX platforms. –M.S. share|improve this answer answered Oct 11 '13 at 8:42 libralhb 11 add a comment| up vote 0 down vote In the first place, str is a pointer that points at "string".

A conformant C implementation could have this program work as expected only on Tuesdays if it wants. :) If you're going to pass a pointer, you have to make sure it

For example, Linux systems using the grsecurity patch may log SIGSEGV signals in order to monitor for possible intrusion attempts using buffer overflows. I'd go with the first option, but it depends on how name_1 is used afterwards. –Jonathan Leffler Nov 27 '12 at 6:51 add a comment| up vote 0 down vote char How can I master those concepts to get rid of segmentation...How can I prevent runtime errors in my program? Scanf Segmentation Fault That's [math]2^{20}[/math] 4KiB segments in all.

When you create a char* initialized to a string, the string data is compiled into the text segment and the program initializes the pointer to point into the text segment. If you have a dire need to strip away the const-ness of the string, then use const_cast. Join them; it only takes a minute: Sign up Segmentation fault - char pointer up vote 4 down vote favorite In the code below, the line: *end = *front; gives a navigate here In your first example, you're getting a pointer to that const data.

This time when the compiler initializes the character array (which is still just a char*) it's pointing into the data segment rather than the text segment, which you can safely alter So, here's a bit of history. char *str is a pointer to a string which is non modifiable(the reason for getting seg fault).. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

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