Why does the strcat function give me a segmentation fault? [duplicate]
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How do I concatenate two strings in C?
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I want to concatenate "/bin/" and "touch" so that I will have "/bin/touch".
In my program, I have
char* filePath = malloc((strlen("/bin/") + strlen(rv[0]))* sizeof(char));
filePath = strcat("/bin/",rv[0])
First of all, rv[0] contains a string, "touch". I allocate 10 bytes in memory by using malloc function, and filePath will be the pointer to those 10 bytes of memory. Because, the total length of the string concatenated ("/bin/touch") will be 10.
The program executes normally until the second line which gives me a segmentation fault. Did I make any mistake on the strcat function?
c segmentation-fault dynamic-memory-allocation
asked Nov 11 at 9:56
ronn Lee
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marked as duplicate by Antti Haapala c
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Nov 11 at 10:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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down vote
favorite
This question already has an answer here:
How do I concatenate two strings in C?
10 answers
How do I concatenate const/literal strings in C?
17 answers
I want to concatenate "/bin/" and "touch" so that I will have "/bin/touch".
In my program, I have
char* filePath = malloc((strlen("/bin/") + strlen(rv[0]))* sizeof(char));
filePath = strcat("/bin/",rv[0])
First of all, rv[0] contains a string, "touch". I allocate 10 bytes in memory by using malloc function, and filePath will be the pointer to those 10 bytes of memory. Because, the total length of the string concatenated ("/bin/touch") will be 10.
The program executes normally until the second line which gives me a segmentation fault. Did I make any mistake on the strcat function?
c segmentation-fault dynamic-memory-allocation
asked Nov 11 at 9:56
ronn Lee
876
marked as duplicate by Antti Haapala c
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Nov 11 at 10:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
You're appending to a string literal.strcatdoes not work like that. You'd strcpy tofilePathand then strcat
– Antti Haapala
Nov 11 at 10:04
1
Also you're not allocating space for the null terminator (+ 1), andsizeof(char)is 1 by definition.
– Antti Haapala
Nov 11 at 10:05
When copying strings,strcat()copies the input string up to and including the zero terminator. Yourmalloc()size does not allow for the presence of that zero terminator, so is one character shorter than needed.strcat()therefore writes past the allocated end offilePath- which gives undefined behaviour. A call ofstrcat()should not even compile when given two string literals. If it is, then either your code is forcing it in some way you haven't shown, or your compiler/library is flawed. I'll bet on the former.
– Peter
Nov 11 at 10:07
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
How do I concatenate two strings in C?
10 answers
How do I concatenate const/literal strings in C?
17 answers
I want to concatenate "/bin/" and "touch" so that I will have "/bin/touch".
In my program, I have
char* filePath = malloc((strlen("/bin/") + strlen(rv[0]))* sizeof(char));
filePath = strcat("/bin/",rv[0])
First of all, rv[0] contains a string, "touch". I allocate 10 bytes in memory by using malloc function, and filePath will be the pointer to those 10 bytes of memory. Because, the total length of the string concatenated ("/bin/touch") will be 10.
The program executes normally until the second line which gives me a segmentation fault. Did I make any mistake on the strcat function?
c segmentation-fault dynamic-memory-allocation
asked Nov 11 at 9:56
ronn Lee
876
This question already has an answer here:
How do I concatenate two strings in C?
10 answers
How do I concatenate const/literal strings in C?
17 answers
I want to concatenate "/bin/" and "touch" so that I will have "/bin/touch".
In my program, I have
char* filePath = malloc((strlen("/bin/") + strlen(rv[0]))* sizeof(char));
filePath = strcat("/bin/",rv[0])
First of all, rv[0] contains a string, "touch". I allocate 10 bytes in memory by using malloc function, and filePath will be the pointer to those 10 bytes of memory. Because, the total length of the string concatenated ("/bin/touch") will be 10.
The program executes normally until the second line which gives me a segmentation fault. Did I make any mistake on the strcat function?
This question already has an answer here:
How do I concatenate two strings in C?
10 answers
How do I concatenate const/literal strings in C?
17 answers
c segmentation-fault dynamic-memory-allocation
c segmentation-fault dynamic-memory-allocation
asked Nov 11 at 9:56
ronn Lee
876
asked Nov 11 at 9:56
ronn Lee
876
asked Nov 11 at 9:56
ronn Lee
876
asked Nov 11 at 9:56
ronn Lee
876
asked Nov 11 at 9:56
ronn Lee
876
876
marked as duplicate by Antti Haapala c
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Nov 11 at 10:06
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marked as duplicate by Antti Haapala c
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Nov 11 at 10:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
You're appending to a string literal.strcatdoes not work like that. You'd strcpy tofilePathand then strcat
– Antti Haapala
Nov 11 at 10:04
1
Also you're not allocating space for the null terminator (+ 1), andsizeof(char)is 1 by definition.
– Antti Haapala
Nov 11 at 10:05
When copying strings,strcat()copies the input string up to and including the zero terminator. Yourmalloc()size does not allow for the presence of that zero terminator, so is one character shorter than needed.strcat()therefore writes past the allocated end offilePath- which gives undefined behaviour. A call ofstrcat()should not even compile when given two string literals. If it is, then either your code is forcing it in some way you haven't shown, or your compiler/library is flawed. I'll bet on the former.
– Peter
Nov 11 at 10:07
add a comment |
You're appending to a string literal.strcatdoes not work like that. You'd strcpy tofilePathand then strcat
– Antti Haapala
Nov 11 at 10:04
1
Also you're not allocating space for the null terminator (+ 1), andsizeof(char)is 1 by definition.
– Antti Haapala
Nov 11 at 10:05
When copying strings,strcat()copies the input string up to and including the zero terminator. Yourmalloc()size does not allow for the presence of that zero terminator, so is one character shorter than needed.strcat()therefore writes past the allocated end offilePath- which gives undefined behaviour. A call ofstrcat()should not even compile when given two string literals. If it is, then either your code is forcing it in some way you haven't shown, or your compiler/library is flawed. I'll bet on the former.
– Peter
Nov 11 at 10:07
You're appending to a string literal.
strcat does not work like that. You'd strcpy to filePath and then strcat– Antti Haapala
Nov 11 at 10:04
You're appending to a string literal.
strcat does not work like that. You'd strcpy to filePath and then strcat– Antti Haapala
Nov 11 at 10:04
1
1
Also you're not allocating space for the null terminator (+ 1), and
sizeof(char) is 1 by definition.– Antti Haapala
Nov 11 at 10:05
Also you're not allocating space for the null terminator (+ 1), and
sizeof(char) is 1 by definition.– Antti Haapala
Nov 11 at 10:05
When copying strings,
strcat() copies the input string up to and including the zero terminator. Your malloc() size does not allow for the presence of that zero terminator, so is one character shorter than needed. strcat() therefore writes past the allocated end of filePath - which gives undefined behaviour. A call of strcat() should not even compile when given two string literals. If it is, then either your code is forcing it in some way you haven't shown, or your compiler/library is flawed. I'll bet on the former.– Peter
Nov 11 at 10:07
When copying strings,
strcat() copies the input string up to and including the zero terminator. Your malloc() size does not allow for the presence of that zero terminator, so is one character shorter than needed. strcat() therefore writes past the allocated end of filePath - which gives undefined behaviour. A call of strcat() should not even compile when given two string literals. If it is, then either your code is forcing it in some way you haven't shown, or your compiler/library is flawed. I'll bet on the former.– Peter
Nov 11 at 10:07
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1 Answer
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Take a look at the reference for how to use strcat:
char *strcat( char *dest, const char *src );
Appends a copy of the null-terminated byte string pointed to by
srcto the end of the null-terminated byte string pointed to bydest.
The first parameter must thus be a pointer to a memory location large enough to hold the bytes of both the C string already there and the C string pointed to by src.
You call strcat("/bin/",rv[0]) and thus try to write into memory where the string literal "/bin/" is stored .. which is normally in readonly memory, thus You get a segmentation fault.
You need to copy "/bin/" first into the allocated memory pointed to by filePath and then append rv[0] there.
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Take a look at the reference for how to use strcat:
char *strcat( char *dest, const char *src );
Appends a copy of the null-terminated byte string pointed to by
srcto the end of the null-terminated byte string pointed to bydest.
The first parameter must thus be a pointer to a memory location large enough to hold the bytes of both the C string already there and the C string pointed to by src.
You call strcat("/bin/",rv[0]) and thus try to write into memory where the string literal "/bin/" is stored .. which is normally in readonly memory, thus You get a segmentation fault.
You need to copy "/bin/" first into the allocated memory pointed to by filePath and then append rv[0] there.
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
add a comment |
up vote
0
down vote
Take a look at the reference for how to use strcat:
char *strcat( char *dest, const char *src );
Appends a copy of the null-terminated byte string pointed to by
srcto the end of the null-terminated byte string pointed to bydest.
The first parameter must thus be a pointer to a memory location large enough to hold the bytes of both the C string already there and the C string pointed to by src.
You call strcat("/bin/",rv[0]) and thus try to write into memory where the string literal "/bin/" is stored .. which is normally in readonly memory, thus You get a segmentation fault.
You need to copy "/bin/" first into the allocated memory pointed to by filePath and then append rv[0] there.
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
add a comment |
up vote
0
down vote
up vote
0
down vote
Take a look at the reference for how to use strcat:
char *strcat( char *dest, const char *src );
Appends a copy of the null-terminated byte string pointed to by
srcto the end of the null-terminated byte string pointed to bydest.
The first parameter must thus be a pointer to a memory location large enough to hold the bytes of both the C string already there and the C string pointed to by src.
You call strcat("/bin/",rv[0]) and thus try to write into memory where the string literal "/bin/" is stored .. which is normally in readonly memory, thus You get a segmentation fault.
You need to copy "/bin/" first into the allocated memory pointed to by filePath and then append rv[0] there.
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
Take a look at the reference for how to use strcat:
char *strcat( char *dest, const char *src );
Appends a copy of the null-terminated byte string pointed to by
srcto the end of the null-terminated byte string pointed to bydest.
The first parameter must thus be a pointer to a memory location large enough to hold the bytes of both the C string already there and the C string pointed to by src.
You call strcat("/bin/",rv[0]) and thus try to write into memory where the string literal "/bin/" is stored .. which is normally in readonly memory, thus You get a segmentation fault.
You need to copy "/bin/" first into the allocated memory pointed to by filePath and then append rv[0] there.
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
answered Nov 11 at 10:06
Daniel Jour
11.8k12046
11.8k12046
add a comment |
add a comment |
You're appending to a string literal.
strcatdoes not work like that. You'd strcpy tofilePathand then strcat– Antti Haapala
Nov 11 at 10:04
1
Also you're not allocating space for the null terminator (+ 1), and
sizeof(char)is 1 by definition.– Antti Haapala
Nov 11 at 10:05
When copying strings,
strcat()copies the input string up to and including the zero terminator. Yourmalloc()size does not allow for the presence of that zero terminator, so is one character shorter than needed.strcat()therefore writes past the allocated end offilePath- which gives undefined behaviour. A call ofstrcat()should not even compile when given two string literals. If it is, then either your code is forcing it in some way you haven't shown, or your compiler/library is flawed. I'll bet on the former.– Peter
Nov 11 at 10:07