c - Strange behaviour of gcc in Debian -


<प्री> char * stft (const char * fmt, ...) {va_list आइटम; चार * बाहर; इंट जादू = 0; // & lt; - यहां जादू? va_start (आइटम, एफएमटी); Vsprintf (बाहर, एफएमटी, आइटम); va_END (आइटम); वापसी; }

जैसे प्रयोग करें: 

  char * str = stft ("% s-% s% s", a, b, c);   
 यह समाधान काम कर रहा है? अगर अप्रयुक्त "जादू" चर को हटा दें - मेरे पास रिटर्न स्ट्रिंग के बाद सेगमेंटेशन फॉल्ट है I क्या गलत है?  
 $ gcc --version जीसीसी (डेबियन 4.4.5-8) 4.4.5  
 $ uname- एक लिनक्स गहरे स्टेशन (निचोड़) 2.6.32- 5-686 # 1 एसएमपी शुक्र 10 मई 08:33:48 यूटीसी 2013 i686 जीएनयू / लिनक्स   
 
 
 आप एक अपरिहार्य संकेतक  बाहर  लिखने की कोशिश कर रहे हैं यही कारण है कि आप दुर्घटना यह बुरी तरह से अपरिभाषित व्यवहार है जादू संयोग है; यह व्यवहार को बेहतर परिभाषित नहीं करता है।  
 इसका उपयोग करना सबसे अच्छा है:  
  char * out = malloc (256); ... vsnprintf (बाहर, 256, एफएमटी, आइटम); ... बाहर निकलो;   
 या ऐसा कुछ।  
 आप इसे सुधार सकते हैं। उदाहरण के लिए:  
  char * stft (const char * fmt, ...) {va_list आइटम; va_start (आइटम, एफएमटी); Int लंबाई = vsnprintf (0, 0, एफएमटी, आइटम); va_end (आइटम); चार * आउट = मॉलोक (लंबाई + 1); अगर (बाहर! = 0) {va_start (आइटम, एफएमटी); Vsnprintf (बाहर, लंबाई + 1, एफएमटी, आइटम); va_end (आइटम); } बाहर निकलना; }   
 सुनिश्चित करें कि आप कॉलिंग कोड में आवंटित स्मृति को मुक्त कर दें।

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