C string operations
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if I have this code : char * s =(char*)malloc(10); and then I write 's="test"' then s will have a different address right ? why is this happening ?
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if I have this code : char * s =(char*)malloc(10); and then I write 's="test"' then s will have a different address right ? why is this happening ?
s is a pointer, when you assign a value to s it should be an address not a literal. You allocate 10 bytes of memory and s holds the address of that memory. To assign a literal to the memory pointed to by s use *s = "test";
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if I have this code : char * s =(char*)malloc(10); and then I write 's="test"' then s will have a different address right ? why is this happening ?
You are allocating assigning the address of the string literal "test" to the pointer
sthus overwriting the address returned bymalloc. You should be usingstrcpyto copy the four characters into the allocated buffer. [edit] better terminology as pointed out by member below. [/edit] -
You are allocating assigning the address of the string literal "test" to the pointer
sthus overwriting the address returned bymalloc. You should be usingstrcpyto copy the four characters into the allocated buffer. [edit] better terminology as pointed out by member below. [/edit]or, even better,
strncpy. -
or, even better,
strncpy. -
if I have this code : char * s =(char*)malloc(10); and then I write 's="test"' then s will have a different address right ? why is this happening ?
Member 10964099 wrote:
why is this happening ?
It append because it designed this way. That is how pointers work. I recommend to read THE book "the C Language" from Kernigan & Ritchie. The C Programming Language - Wikipedia, the free encyclopedia[^]
Patrice “Everything should be made as simple as possible, but no simpler.” Albert Einstein
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if I have this code : char * s =(char*)malloc(10); and then I write 's="test"' then s will have a different address right ? why is this happening ?
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if I have this code : char * s =(char*)malloc(10); and then I write 's="test"' then s will have a different address right ? why is this happening ?
What all the others said and more generally rarely if ever do use an equal sign on a string in C at any other time than creation. There are functions required to add, copy, compare, split and search strings and you can't just use mathematical symbols like "=" and "+" to do things with strings like you can in some higher languages. Strings are not mathematical types and you can't use mathematical functions to do things with them. Those coming from a coding background in Java or Basic tend to do what you did and you have to unlearn it when using C because "string1" + "string2" will not work either, you will need to use strcat functions. String compares are likewise a function and you can't use mathematical equality signs to compare two strings. Learning the string functions, it is a requirement of learning the C language.
In vino veritas
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You are allocating assigning the address of the string literal "test" to the pointer
sthus overwriting the address returned bymalloc. You should be usingstrcpyto copy the four characters into the allocated buffer. [edit] better terminology as pointed out by member below. [/edit]Shouldn't that be: "you are assigning the address" instead of allocating?
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Shouldn't that be: "you are assigning the address" instead of allocating?