I am practicing printf formatting options. I want to print out a
string literal as a raw string as opposed to its special meaning. For
e.g I want to print out the "%f" part in:
printf("%f\n");
literally as "%f" rather than a zero value floating point/constant
double (0.000000).
I tried to escape the % sign but the compiler complained with a
warning about an illegal escape sequence.
I am using the Visual C++ 6.0 compiler.
A one-off solution could be to print it out as a string literal as:
char* s = 0;
s = "%f";
printf("%s\n", s);
But that'd work only for this format. I want a flexible solution since
I want to print out many of these formats as literals such as:
"%2.8f"
"%14.3f"
"%.9f"
etc.
4  8161 
On Jun 16, 3:31 pm, Mukesh_Singh_N. ..@yahoo.com wrote:
I am practicing printf formatting options. I want to print out a
string literal as a raw string as opposed to its special meaning. For
e.g I want to print out the "%f" part in:
printf("%f\n");
literally as "%f" rather than a zero value floating point/constant
double (0.000000).
I tried to escape the % sign but the compiler complained with a
warning about an illegal escape sequence.
I am using the Visual C++ 6.0 compiler.
A one-off solution could be to print it out as a string literal as:
char* s = 0;
s = "%f";
printf("%s\n", s);
But that'd work only for this format. I want a flexible solution since
I want to print out many of these formats as literals such as:
"%2.8f"
"%14.3f"
"%.9f"
etc.
For instance, I could also use:
puts, or
putch, or
putchar, or
putche
but I want to only use printf because I am practicing. Suggestions,
please. Mu************* **@yahoo.com wrote:
I am practicing printf formatting options. I want to print out a
string literal as a raw string as opposed to its special meaning. For
e.g I want to print out the "%f" part in:
printf("%f\n");
printf("%%f\n") ;
literally as "%f" rather than a zero value floating point/constant
double (0.000000).
I tried to escape the % sign but the compiler complained with a
warning about an illegal escape sequence.
I am using the Visual C++ 6.0 compiler.
A one-off solution could be to print it out as a string literal as:
char* s = 0;
s = "%f";
printf("%s\n", s);
But that'd work only for this format. I want a flexible solution since
I want to print out many of these formats as literals such as:
"%2.8f"
"%14.3f"
"%.9f"
printf("%s\n", "%2.8f");
printf("%s\n", "%14.3f");
printf("%s\n", "%.9f");
would all work.
On Jun 16, 3:36 pm, Harald van D k <true...@gmail. comwrote:
Mukesh_Singh_N. ..@yahoo.com wrote:
I am practicing printf formatting options. I want to print out a
string literal as a raw string as opposed to its special meaning. For
e.g I want to print out the "%f" part in:
printf("%f\n");
printf("%%f\n") ;
literally as "%f" rather than a zero value floating point/constant
double (0.000000).
I tried to escape the % sign but the compiler complained with a
warning about an illegal escape sequence.
I am using the Visual C++ 6.0 compiler.
A one-off solution could be to print it out as a string literal as:
char* s = 0;
s = "%f";
printf("%s\n", s);
But that'd work only for this format. I want a flexible solution since
I want to print out many of these formats as literals such as:
"%2.8f"
"%14.3f"
"%.9f"
printf("%s\n", "%2.8f");
printf("%s\n", "%14.3f");
printf("%s\n", "%.9f");
would all work.- Hide quoted text -
- Show quoted text -
Thanks very much, friend. I almost forgot about the "%%" format
specifier. I had seen a C reference page mention it long ago. I can't
find it now. Can you please point me to a reference page that explains
it?
On Jun 16, 3:45 pm, Mukesh_Singh_N. ..@yahoo.com wrote:
On Jun 16, 3:36 pm, Harald van D k <true...@gmail. comwrote:
Mukesh_Singh_N. ..@yahoo.com wrote:
I am practicing printf formatting options. I want to print out a
string literal as a raw string as opposed to its special meaning. For
e.g I want to print out the "%f" part in:
printf("%f\n");
printf("%%f\n") ;
literally as "%f" rather than a zero value floating point/constant
double (0.000000).
I tried to escape the % sign but the compiler complained with a
warning about an illegal escape sequence.
I am using the Visual C++ 6.0 compiler.
A one-off solution could be to print it out as a string literal as:
char* s = 0;
s = "%f";
printf("%s\n", s);
But that'd work only for this format. I want a flexible solution since
I want to print out many of these formats as literals such as:
"%2.8f"
"%14.3f"
"%.9f"
printf("%s\n", "%2.8f");
printf("%s\n", "%14.3f");
printf("%s\n", "%.9f");
would all work.- Hide quoted text -
- Show quoted text -
Thanks very much, friend. I almost forgot about the "%%" format
specifier. I had seen a C reference page mention it long ago. I can't
find it now. Can you please point me to a reference page that explains
it?- Hide quoted text -
- Show quoted text -
OK, I got it. http://www.cppreference.com/stdio/printf.html
Thanks very much again. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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