Why does Graphviz fail on gvLayout?
- by David Brown
Once again, here I am writing C without really knowing what I'm doing...
I've slapped together a simple function that I can call from a C# program that takes a DOT string, an output format, and a file name and renders a graph using Graphviz.
#include "types.h"
#include "graph.h"
#include "gvc.h"
#define FUNC_EXPORT __declspec(dllexport)
// Return codes
#define GVUTIL_SUCCESS 0
#define GVUTIL_ERROR_GVC 1
#define GVUTIL_ERROR_DOT 2
#define GVUTIL_ERROR_LAYOUT 3
#define GVUTIL_ERROR_RENDER 4
FUNC_EXPORT int RenderDot(char * dotData, const char * format,
const char * fileName) {
Agraph_t * g; // The graph
GVC_t * gvc; // The Graphviz context
int result; // Result of layout and render operations
// Create a new graphviz context
gvc = gvContext();
if (!gvc) return GVUTIL_ERROR_GVC;
// Read the DOT data into the graph
g = agmemread(dotData);
if (!g) return GVUTIL_ERROR_DOT;
// Layout the graph
result = gvLayout(gvc, g, "dot");
if (result) return GVUTIL_ERROR_LAYOUT;
// Render the graph
result = gvRenderFilename(gvc, g, format, fileName);
if (result) return GVUTIL_ERROR_RENDER;
// Free the layout
gvFreeLayout(gvc, g);
// Close the graph
agclose(g);
// Free the graphviz context
gvFreeContext(gvc);
return GVUTIL_SUCCESS;
}
It compiles fine, but when I call it, I get GVUTIL_ERROR_LAYOUT. At first, I thought it might have been how I was declaring my P/Invoke signature, so I tested it from a C program instead, but it still failed in the same way.
RenderDot("digraph graphname { a -> b -> c; }", "png", "C:\testgraph.png");
Did I miss something?
EDIT
If there's a chance it has to do with how I'm compiling the code, here's the command I'm using:
cl gvutil.c /I "C:\Program Files (x86)\Graphviz2.26\include\graphviz"
/LD /link /LIBPATH:"C:\Program Files (x86)\Graphviz2.26\lib\release"
gvc.lib graph.lib cdt.lib pathplan.lib
I've been following this tutorial that explains how to use Graphviz as a library, so I linked to the .lib files that it listed.
